ChirONq ill
PHlTdNG TRINH VA HE PHl/dNG TRINH
P h a n 1 . EOrolllirO VAIXr D E CUA C H I / O R T G L
NOIDUNG Ndi dung chfnh cua chuong gom:
Phuong trinh bae nha't : Giai va bien luan phuong trinh bae nhat cd chiia tham so, phuong trinh quy vl phuong trinh bae nhit. - Phuang trinh bae hai : Giai va bien luan phuong trinh bae hai, dinh If Vi-et, mdt so phuong trinh quy vl phuang trinh bae hai. He phuong trinh bae nhit va he phuong trinh bae hai. II.
MUC Tl£U
1.
Kien thiirc
Hieu khai niem phuong trinh, phuong trinh tuong duong, phuang trinh ha qua; bia't dugc cac phep bien doi tuong duong va phep bia'n doi cho phuong trinh ha qua. Nim viing cdng thiic va cac phuong phap giai phuong trinh bae nha't, phuong trinh bae hai mdt in va he phuong trinh bae nhat, bae hai 2 in. Hieu y nghia hinh hgc ciia cac nghiem ciia phuong trinh va he phuong trinh bae nha't va bae hai. 2.
KT nang Bia't each giai va bien luan : + Phuong trinh bae nha't va bae hai mdt in, + Phuang trinh dang |ax + b| = [ex + d| va phuong trinh chiia in d miu, 235
+ Phuang trinh trung phuang + He hai phuong trinh bae nha't 2 in (bing dinh thiic cip hai) • Bia't each giai (khdng bien luan): + He ba phuang trinh bae nhit ha in, + He phuang trinh bae hai Bia't giai mdt so bai toan vl tuang giao giiia dd thi cua hai ham sd bae khdng qua 2. 3.
Thai do - HS cd tfnh cin than, kidn tri va khoa hgc khi tim giao cua hai d6 thi.
HS tha'y dugc quan he mat thiet gifla toan hgc va ddi sdng, toan hoc xua't hiendo nhu ciu tur ddi sdng.
236
P h a n a . CAC B A I SOAJV
§1. Dai cvtdng ve phtfofng trinh (tiet 1, 2) L
MUC TifiU
1.
Kie'n thurc Giiip HS :
Hiiu khai niem phuang trinh, tap xac dinh (diiu kien xac dinh) va tap nghiem ciia phuong trinh. - Hia'u khai niem phudng trinh tuong duong va cac phep bien doi tuong duong. 2.
KT nang
Biet each thii xem mdt so cho trudc cd phai la nghiem ciia phuang trinh da cho hay khdng. Bia't sii dung cac phep bia'n ddi tuong duong thudng diing. 3.
Thai do Ren luyen tfnh nghiam tiic khoa hgc.
II.
CHUAN BI CUA GV VA HS
1.
Chuan bi ciia GV : Chuin bi bai ki cac kia'n thiic ma HS da hgc d ldp 9 de dat cau hdi. Chuin bi mdt sd hinh ve trong SGK; pha'n mau,...
2.
Chuan hi ciia HS : Cin dn lai mdt so kie'n thiic vl ham sd da hgc d ldp 9.
ra.
PHAN P H 6 I T H 6 I
LUONG
Bai nay chia lam 2 tia't: Tie't thut nhdt: tic ddu de'n he't phdn 3; Tiet thvc hai: ta phdn 4 den het phdn bdi tap.
237
IV.
TIEN TRINH DAY HOC
A. Bai cii c a u hdi 1 Hay cho bia't nghiem ciia phuong trinh 2x - 1 = x + 4 c a u hdi 2 Cac sd nao sau day la nghiem cua phuong trinh x^ + x = V2x, 1;2;-1;0
B. Bai mdi HOATDONGI 1.
Khai niem phuang trinh mot an Dinh nghTa. Cho hai ham sd y = f(x) va y = g(x) cd tap xac dinh lin lugt la
% va 2)g. Dat 3) = 3)f n 2)^. Menh dl chiia bia'n f(x) = g(x) dugc ggi la mdt phuang trinh mdt in; X ggi la dn 50'(hay dn) va 3) ggi la tap xdc dinh cua phuong trinh. Sd XQ e S) dugc ggi la mdt nghiem ciia phuang trinh f(x) = g(x) ne'u f(Xo) = g(Xo) la menh dl diing. GV: Thuc hien thao tdc ndy tron^ 3 phut. Hoat ddng cua GV
Hoat ddng ciia HS
c a u hdi 1 Cy'.fi y tra Idi cau hdi 1 Hay neu mdt vf du vl phuang Chang ban Vx +1 = x + 1 . trinh mot in. Ggi y tra Idi cau hdi 2 c a u hdi 2 Tap xac dinh ciia phuang trinh nay Hay nau tap xac dinh cua la [1; + 00). phuang trinh vira nau
238
Ggi y tra Idi cau hdi 3 Cau hdi 3 Hay chi ra mdt nghiem ciia Chang ban x = 1 la nghiem. phuang trinh. Chu y 1 1) De thuan tien trong thuc hanh, ta khdng cin via't rd tap xac dinh ii) cua phuong trinh ma chi cin neu diiu kien de x e 3). Diiu kien dd ggi la diiu kien xac dinh ciia phuang trinh, ggi tit la diiu kien ciia phuang trinh. Nhu vay, diiu kien ciia phuong trinh bao gom cac diiu kien de gia tri cu.a f(x) va g(x) ciing dugc xac dinh va cac diiu kien khac ciia in (ne'u cd yau cau) (theo quy udc vl tap xac dinh ciia ham sd cho bdi bieu thiic). GV: Thuc hien thao tdc ndy trong 3 phdt. Hoat ddng ciia HS
Hoat ddng ciia GV c a u hdi 1
Ggi y tra Idi cau hdi 1
Hay ndu mdt thuat ngii khac Diiu kien xac^nh ciia phuong trinh ve tap xac dinh cua phuang hoac diiu kien cua phuang trinh. trinh. c a u hdi 2
Ggi y tra Idi cau hdi 2 Hay neu md'i quan he giiia tap nghiam va tap xac dinh cua Tap nghiam la tap con cua tap xac dinh cua phuang trinh. phuang trinh.
Vi du 1. a) Diiu kien xac dinh ciia phuong trinh V x ^ - 2 x ^ + 1 = 3 la x^ - 2x^ + 1 > 0. b) Khi tim nghiem nguyan cua j)huong trinh 2
= Vx ta hieu diiu X
kian ciia phuang trinh l a x e Z,X7tOvaj[:>0 (hay x nguyan duong).
239
Chu y 2. 1) Khi giai mdt phuang trinh (tiic la tim tap nghiem ciia phuang trinh), nhilu khi ta chi cin, hoac chi cd the tfnh gia tri gin dung cua nghiam (vdi do chinh xac nao dd). Gia tri dd ggi la nghiem gin dung cua phuang trinh. Ching ban, bing may tfnh bd tiii, ta tfnh nghiem gin diing (chfnh xac da'n hang phin nghin) ciia phuong trinh x = 7 la x « 1,913. 2) Nghiem ciia phuang trinh f(x) = g(x) chfnh la hoanh do cac giao diem cua do thi hai ham sd y = f(x) va y = g(x). GV: Chi neu rdt nhanh cdc nhdn xet ciia chd y tren.
HOATDONG2
2.
Phuong trinh tuong duang Ta da bia't : hai phuang trinh tuang duang na'u chiing cd cung mot tap
nghiem. Neu phuang trinh fi(x) = gi(x) tuong duong vdi phuang trinh f2(x) = g2(x) thi ta via't: fi(x) = gi(x) ^
f2(x) = g2(x).
GV: Nhdn mqnh : - Hai phuang trinh ma tap nghiam ciia phuang trinh nay bang tap nghiem ciia phuong trinh kia thi tuong duong vdi nhau. - Hai phuong trinh cung vd nghiem thi tuong duong. GV: Hudng ddn HS thuc Men|H1| vd thuc Men thao tdc ndy trong 5 phdt. Hoat ddng cua GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Tim tap nghiem cua phuang Tap nghiem ciia phuofng trinh nay la: trinh : S={1} Vx-l=2Vl-x 240
c a u hdi 2
Ggi y tra Idi cau hdi 2 Tim tap nghiem cua phuang Tap nghiam ciia phuong trinh nay la trinh : {1} X - 1 = 0.
c a u hdi 3 Ggi y tra Idi cau hdi 3 Hai phuang trinh nay cd Hai phuang trinh nay tuang duang, tuang duang khdng? vi sao? vi chiing cd ciing tap nghiem. Cau hdi 4 Ggi y tra Idi cau hdi 4 Tim tap nghiem cua phuang Tap nghiem cua phuong trinh nay la 0 trinh X + Vx-2 = 1 +Vx-2
Ggi y tra Idi cau hdi 5 c a u hdi 5 Tap nghiam cua phuang trinh nay la Tim tap nghiem ciia phuang {1} trinh X - 1 = 0. c a u hdi 6 Ggi y tra Idi cau hdi 6 Hai phuang trinh nay cd Hai phuang trinh nay khdng tuang tuang duang khdng? vi sao? duang, vi chiing khdng cd ciing tap nghiem. c a u hdi 7 Ggi y tra Idi cau hdi 7 Tim tap nghiem ciia phuang Tap nghiam ciia phuang trinh nay la trinh {-i;i} 1x1 = 1 Cau hdi 8 Ggi y tra Idi cau hdi 8 Tim tap nghiem ciia phuang Tap nghiem ciia phuong trinh nay la trinh {1} x = 1.
le-TKBGBAISOIONC-TI
241
c a u hdi 9 Ggi y tra Idi cau hdi 9 Hai phuang trinh nay cd Hai phuang trinh nay khdng tuang tuang duang khdng? vi sao? duang, vi chiing khdng cd ciing tap nghiem. GV: Neu cdc nhdn xet sau : Khi mudn nha'n manh hai phuong trinh cd ciing tap xac dinh ^ (hay co ciing diiu kien xac dinh ma ta ciing kf hieu la 9^) va tuong duong vdi nhau, ta ndi : - Hai phuong trinh la tuong duong vdi nhau tren ID, hoac - Vdi dieu kien if*, hai phuong trinh la tuong duong vdi nhau. Chang ban: Ta ndi, vdi x > 0, hai phuong trinh x = 1 va x = 1 la tuong duong vdi nhau. Trong cac phep bia'n doi phuong trinh, dang chii y nhat la cac phep bien doi khdng lam thay doi tap nghiem ciia phuang trinh. Ta ggi chiing la cac phep bia'n doi tuong duang. Phep Men ddi tuang duang Men mot phuang trinh thdnh phuang trinh tuang duang vdi nd. Dudi day la dinh If vl cac phep bien doi tuong duong thudng diing. Dinh Ii 1. Cho phuang trinh f(x) = g(x) cd tap xdc dinh iP; y = h(x) la mot hdm sd xdc dinh tren iD (h(x) cd the Id mot hang sd). Khi dd tren y\ phuang trinh dd cho tuang duang vdi mdi phuang trinh sau : \)f(x) + h(x) ^g(x) + h(x); 2)f(x) h(x) = g(x) h(x) ne'u h(x) ^ 0 vdi mgi A: e 2). GV: Hudng ddn HS chicng minh nhanh dinh li. Tit dinh If tran, ta dl thay : hai quy tic bien doi phuong trinh da hgc d ldp dudi (quy tic chuyen ve va quy tic nhan vdi mdt sd khac 0) la nhiing phep bien doi tuong duong.
242
GV: Hudng ddn HS thuc hien\H2\ vd thuc hien thao tdc ndy trong S phut. Hoat ddng ciia HS
Hoat ddng cua GV Cau hdi 1
Ggi y tra Idi cau hdi 1 O cau a), sau khi chuyen ve' Cd. cd dugc phuang trinh tuang Theo dinh If tren. duang hay khdng? Ggi y tra Idi cau hdi 2 Sau khi luge bd ta dugc phuang b) Cho phuang trinh trinh 3x = x^ Phuang trinh nay cd 3x + V x - 2 = x^ + V x - 2 . hai nghiam x = 0 va x = 3. nhung Luge bd V x - 2 d ca hai ve X = 0 khdng phai la nghidm ciia ciia phuang trinh thi dugc phuang trinh ban dau. phuang trinh tuang duang. Hai phuang trinh nay khdng tuang duang.
cau hdi 2
HOAT DONG 3 Phuong trinh he qua GV: Neu vi du 2 de ddt vdn de ve phuang trinh he qua. Xet phuong trinh : Vx = 2 - X.
(1)
Binh phuong hai ve, ta dugc phuong trinh mdi : X = 4 - 4x + x^
(2)
Tap nghiem ciia (1) la Sj = {1}, cua (2) la S2 ={ 1; 4}. Hai phuong trinh (1) va (2) khdng tuong duong. Tuy nhian ta thay S2 3 Sp Trong trudng hop nay ta ggi (2) la phuong trinh he qua ciia phuong trinh (1). Tong quat, fi(x) = gi(x) goi la phuang trinh he qua ciia phuong trinh f(x) = g(x) neu tap nghiem cua nd chita tap nghiem cua phuong trinh f(x) = g(x). Khi dd ta viet f(x) = g(x)=^fi(x) = g,(x). 243
GV: Thuc hien thao tdc ndy trong 2 phiu. Hoat ddng cua GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Hay ndu vf du ve hai phuang x = l = > x ^ = l . trinh he qua:
1
c a u hdi 2 Ggi y tra Idi cau hdi 2 Hay chi ra nghiem ngoai lai. Nghiem ngoai lai cua phugng trinh (nghiem ngoai lai ciia phuang la-1." trinh la nhiing nghiem ciia phuang trinh he qua ma khdng phai la nghiem cua phuang trinh ban diu) Tii dinh nghia nay, ta suy ra : Neu hai phuang trinh tuang duang thi mdi phuang trinh deu la he qua ciia phuang trinh con lai. Trong vf du 2, gia tri X = 4 la nghiem cua (2) nhung khdng la nghiem ciia (1). Nan X = 4 la nghiem ngoai lai cua phuang trinh (1). GV: Huang ddn HS thuc hien |H3| vd thuc hien thao tdc ndy trong 2 phiit. Hoat ddng ciia GV
Hoat ddng ciia HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Khing dinh sau day diing hay Diing vi hai phuong trinh nay tuong sai? duong. a) V x - 2 = 1 = > x - 2 = 1? Ggi y tra Idi cau hdi 2 c a u hdi 2 Khang dinh sau day diing Diing, vi phuang trinh diu vd nghiem con phuang trinh sau cd hay sai? nghiem x = 1. , X x(x-l)
b) -5^ ^ = l = > x = 1? x-1 Trong cac phep bie'n doi chi cho phuang trinh he qua, dang chu y la phep bia'n doi dugc nau trong dinh If sau. 244
Djnh II 2. Khi binh phuong hai ve cua mdt phuong trinh, ta dugc phuong trinh he qua cua phuong trinh da cho. Ndi each khac : f(x) = g ( x ) ^ [ f ( x ) f = [g(x)f GV: Thuc hien thao tdc ndy trong 5 phut. Hoat ddng cua HS
Hoat ddng cua GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 Hay ndu mot vi du. ap dung Vx +1 = 2x +1 ^ X +1 = f 2x +1 j ^ dinh If 2. c a u hdi 2 Ggi y tra Idi cau hdi 2 Hay chiing to day la phep Phuang trinh dau chi cd nghiem bia'n doi he qua. X = 0; nhung phuang trinh sau cd 3 hai nghiam x = Ova x = 4 Chii y 1) Cd the chiing minh ring na'u hai va' ciia mdt phuang trinh ludn cung ddu thi khi binh phuong hai ve' cua nd, ta dugc phuong trinh tuong duong. 2) Na'u phep bien doi mdt phuong trinh din den phuong trinh he qua thi sau khi giai phuong trinh he qua, ta phai thit Iqi eae nghiem tim dugc vao phuong trinh da cho de phat hien va loai bd nghiem ngoai lai. GV: Thuc Men thao tdc ndy trong 2 phiit. Hoat ddng ciia GV
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 c a u hdi 1 Hay dat diiu kien cho x de khi binh phuong hai va' ciia phuong Diiu kien x > — 2 trinh Vx - 1 = 2x +1 ta dugc phuong trinh tuong duong. c a u hdi 2 Tim nghiem eiia phuong trinh.
Ggi y tra Idi cau hdi 2 Binh phuang hai ve' ta dugc : 245
3 4x^ + 3x = 0 o X = 0 va x = - - Kiem 4 tra dieu kien ta duoc x = 0 la nghiem. GV: Neu vd hitdng ddn HS gidi vi du 3 trong SGK hoac cd the thay baitg vi du khdc tuang tie HOATDONG 4 4.
Phuong trinh nhieu an
Trong thuc te, ta con gap cac phuong trinh cd nhilu ban mot in. Dd la cac phuong trinh dang F = G, trong dd F va G la cac bieu thiic cua nhilu bie'n. Vi du : 2x + 4xy - y^ = -X + 2y + 3
(3)
la mdt phuang trinh hai dn (x va y); X + y + z = 3xyz
(4)
la mdt phuang trinh ba an (x, y va z). Neu phuong trinh hai in x va y trd thanh menh dl diing khi x = Xy va y = y,, (vdi Xo va y„ la cac sd) thi ta ggi cap sd (x^{, y,,) la mdt nghiem ciia phuong trinh. Ching ban, cap sd (1; 0) la mdt nghiem ciia phuong trinh (3). Khai niem nghiem ciia phuong trinh ba in, bd'n an,... ciing dugc hieu tuong tu. Ching han bd ba sd (1; 1; 1) la mdt nghiem cua phuong trinh (4). Ddi vdi phuong trinh nhieu in, cac khai niem : tap xac dinh (diiu kien xac dinh), tap nghiem, phuong trinh tuong duong, phuong trinh he qua,... ciing tuong tu nhu ddi vdi phuong trinh mdt in. GV: Thitc hien thao tdc ndy trong 5' Hoat ddng cua GV
Hoat ddng ciia HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Hay neu mot vi du ve phuang X + y = 5. trinh 2 an. c a u hdi 2 Hay chi ra mdt nghiem ciia nd. 246
Ggi y tra Idi cau hdi 2 (0;5),(1;4),...
c a u hdi 3 Ggi y tra Idi cau hdi 3 Hay ndu mdt vf du ve phuang X + y = xy. trinh 3 in. c a u hdi 4 Ggi y tra Idi cau hdi 4 Hay chi ra mot nghiem cua (0; 1),(0;2),(1;0),... phuang trinh. HOATDONG 5 5.
Phuong trinh chiia tham sd
Chung ta con xet ca nhiing phuong trinh, trong do ngoai in x con cd nhiing chii khac. Cac chii nay dugc xem la nhiing sd cho trudc va dugc ggi la tham sd. Ching ban, phuong trinh m(x + 2) = 3mx - 1 la mdt phuong trinh chiia tham sd m. GV: Hudng ddn HS thuc hien\\\'5\ vd thuc Men thao tdc ndy trong 5 phiit. Hoat ddng cua GV
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 cau hdi 1 Khi m = 0, hay chi ra phuang 2 = 1 ; Tap nghiem la 0 . trinh va tap nghiem. cau hdi 2 Ggi y tra Idi cau hdi 2 Hay chi ra mdt nghiem phuang trinh khi m ^ 0. m
r"-'i.
Rd rang nghiem va tap nghiem ciia mdt phuang trinh chiia tham so phu thudc vao tham so. Khi giai phuang trinh chiia tham so, ta phai chi ra tap nghiem ciia phuang trinh trong mdi trudng hgp cd the cua tham sd. De nhain manh y dd khi giai phuang trinh chiia tham so, ta thudng ndi la gidi vd Men ludn phuang trinh.
247
TOM TAT BAI HOC 1.
Cho hai ham sd y = f(x) va y = g(x) cd tap xac dinh lin lugt la iDf va 9")^. Daty) = ii)f n
^\.
Menh dl chiia bien f(x) = g(x) dugc ggi la mdt phuong trinh mdt in; X ggi la in sd (hay in) va 9) ggi la tap xac dinh ciia phuang trinh. Sd XQ e ^ ggi la mdt nghiem ciia phuong trinh f(x) = g(x) na'u f(Xo) = g(Xo) la menh di dung. 2.
Hai phuong trinh tuong duong ne'u chiing cd cung mdt tap nghiem. Neu phuong trinh fi(x) = gi(x) tuong duong vdi phuang trinh f2(x) = g2(x) thi ta via't: fi(x) = gi(x)c^ f2(x) = g2(x).
3.
Dinh li 1. Cho phuong trinh f(x) = g(x) cd tap xac dinh iT); y = h(x) la mdt ham sd xac dinh trdn 9) (h(x) cd the la mdt hing so). Khi do trdn ^, phuong trinh da cho tuong duang vdi mdi phuang trinh sau : l)f(x) + h(x) =g(x) + h(x); 2) f(x) h(x) = g(x) h(x) neu h(x) 9^ 0 vdi mgi x e ^J).
4.
fj(x) = gi(x) ggi la phuong trinh he qua ciia phuang trinh f(x) = g(x) neu tap nghiem ciia nd chiia tap nghiem ciia phuong trinh f(x) = g(x). Khi do ta viet f(x) = g(x)=>fi(x) = gi(x).
5.
Dinhli2. Khi binh phuong hai vi ciia mdt phuong trinh, ta dugc phuong trinh he qua cua phuong trinh da cho. Ndi each khac : f(x) = g ( x ) ^ [ f ( x ) f = [g(x)]^
248
HUdNG
DAN T R A LCJI C A U
HOI vA
BAI T A P
SGK
Bail. GV: Hudng ddn cdu a) Hoat ddng cua HS
Hoat ddng ciia GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 Tim dieu kien xac dinh ciia fx^O <^ =:>X = 0 phuang trinh. [-x>0 c a u hdi 2 Ggi f tra Idi cau hdi 2 Tim tap nghiem ciia phuang {0}. trinh.
Trd Idi: Dieu kien ciia phuang trinh
Tap nghiem
a) Vx = V-x
x=0
{0}
b) 3 x - V x - 2 = V 2 - x + 6
x=2
{2}
x>3, x<3vax;t3 Khdng cd x nao
0
X > 1 va X < 0,
0
V3-X c)
1 r =x+ vx-3
x-3 d) X + Vx - 1 = V-x
Khdng cd x nao
Bai 2. GV: Hudng ddn cdu a) Hoat ddng ciia GV
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 c a u hdi 1 Tim diiu kien xac dinh cua X > 1 . phuang trinh. Ggi y tra Idi cau hdi 2 c a u hoi 2 Tim tap nghiem cua phuang Phuang trinh vd nghiem. Tap nghiem 0 . trinh.
249
Trd Idi: b) Vdi dieu kien x > 1 ta cd : x + V x - 1 =0,5 + Vx-1 <=> X = 0,5 (loai vi khdng thoa man diiu kien x
> !)•
Vay phuong trinh vd nghiem. c) X = 6; d) Vd nghiem.
Bai 3. GV: Hudng ddn cdu a) Hoat ddng ciia GV
Hoat ddng ciia HS
Cau hdi 1 Ggi y tra Idi cau hdi 1 Tim diiu kian xac dinh ciia x # l . phuang trinh. c a u hdi 2 Ggi y tra Idi cau hdi 2 Tim tap nghiem ciia phuang Vdi diiu kien x ;^ 1, ta cd : trinh 1 2x-l x+ = x-1 x-1 <^
<:^
jc^-jc+l=2x-l
'x = l _x = 2
Dd'i chieu vdi diiu kien ta ket luan: phuang trinh cd nghiem x =2. Trd Idi: lb) Vdi dieu kien x ^t 2, ta cd :
250
x+- i _ ^ 2 x _ 3 ^ x-2 x-2
x^-2x+l=2x-3
c^ X = 2 (loai do diiu kien x T^ 2) Vay phuong trinh vd nghiem. c) Dieu kien : x > 3. Da tha'y X = 3 la mdt nghiem. Neu X > 3 thi X - 3 ^ 0. Do dd : (x^ - 3x + 2) V x ^ = 0 o
x^-3x + 2 = 0
<» X = 1 hoac X = 2 (hai gia tri diu bi loai do diiu kian x > 3). Vay phuong trinh cd mdt nghiem x = 3 d)xe{-l;2}. Chu y : Binh phuong hai v6 ta chi thu dugc phuong trinh he qua nan phai thii lai de ket luan nghiem.
Bai 4. GV : Hudng ddn cdu a) Hoat ddng ciia GV
Hoat ddng cua HS
cau hdi 1
Ggi y tra Idi cau hdi 1 Tim dieu kien xac dinh ciia 3 < A' < 4,5 phuang trinh.
Ggi y tra Idi cau hdi 2 c a u hdi 2 V x - 3 = V9-2x = > x - 3 = 9 Tim tap nghiem cua phuang - 2x ^ X = 4. trinh. Thii lai thay diing. Vay phuong trinh cd nghiem jc = 4. Trd Idi: b) Ta cd :
251
Vx-l = x - 3 =» X - 1 = (x - 3)^ => x^ - 7x + 10 = 0 => X = 2 hoac X = 5. Thii lai, gia tri x = 2 khdng thoa man. Vay phuong trinh cd nghiem x = 5. c) Ta cd : 2| X - 1 1 = X + 2 ^ 4(x -1)^ = (X + 2f ^ 3x^ -12x = 0 => X = 0 hoac X = 4. Thii lai thay ca hai diu nghiem diing. Vay phuang trinh cd hai nghiem x = 0 va x = 4. d) Ta cd : 1 X - 2 1 = 2x - 1 ^ (x - 2)^ = (2x - 1)^ =^ 3x^ = 3 =>x = ± 1. Thii lai, ta thay chi cd x = 1 nghiem ddng. Vay phuong trinh cd nghiem x = 1.
252
MOT SO B A I T A P T R A C NGHIEM 1.
Cho phuang trinh : x^2 + 1- =
1
Vx^'
Diiu kien ciia phuong trinh la (a)R;
(b)xGR, x > l ;
(c)xeR, x>l;
(d)xeR, x ^ l .
Hay chgn ket qua diing. 2.
Phuong trinh |x| + 1 = x +Vx . Trong cac sd sau day so nao la nghiem cua phuong trinh
3.
(a)-2;
(b)-l;
(c)l;
(d)0.
Trong cac phuang trinh sau, phuong trinh nao tuong duong vdi phuong trinh x^ = 1
4.
(1)
(a) x^ + 3x - 4 = 0;
(b) x^ - 3x - 4 = 0;
(c)|x|=l;
( d ) x ^ + V x = 1+Vx
Cho phuong trinh x^ + x + V x + l = 0 .
(1)
Hay diin diing - sai vao cac ket qua sau day: (a)(l)<::>x^ + x + l = - V x
DDiing
QSai
(b)(l)c^x^ + x + 1 + V x ^ =-Vx + V x ^
DDiing
DSai
(c)(l)<::>x+1 + — + - = 0
Doling
DSai
DDiing
DSai
X
(d)(1) <:^x^ = - l 5.
X
Cho phuong trinh Vx + 1 = x . ^
(1) 253
Hay chgn diing - sai trong cac khang dinh sau ( a ) ( l ) o x + 1 =x^
DDung
DSai
(b)(1) < ^ x + 1 =x^
DDung
DSai
(c) (1) <^ Vx + 1 + Vx = x^ + Vx
DDiing
DSai
(d)(1) <=> VVxTT =Vx
DDiing DSai.
Cho phuong trinh x^ + ( m - l)x + m - 2 = 0
(1)
Hay chgn ket luan diing trong cac ket luan sau (a) phuong trinh (1) vd nghiem Vm; (b) phuong trinh (1) cd 3 nghiem Vm; (c) phuong trinh (1) cd 2 nghiem la x = - 1 va x = 2 - m; (d) ca ba ket luan tren diu sai. Cho phuong trinh X+
1 =1 x+ 1
(1)
(a) phuong trinh (1) cd 2 nghiem la x = - 1 va x = 0; (b) phuong trinh (1) cd 2 nghiem la x = 2 va x = 0; (c) phuong trinh (1) cd 2 nghiem la x = 1 va x = 2; (d) ca ba ket luan tran diu sai. Hay chgn khing dinh diing.
Ddp dn:
254
l-(b)
2. (b)
3. (c)
4.(a) D,
(b)D,
(c)D,
(d)D
5. (a) Sai;
(b) Diing;
(c) Sai;
(d) Diing
6. (e).
7. (d).
BAITAPTUGIAI 8.
Cho phuong trinh x^ + 2005x + VxTI = 1
(1)
Hay chgn ket luan dung trong cac ket luan sau. (a) X = 1 va X = 5 la nghiem cua (1); (b) X = 1 la nghiem ciia (1); (c) X = 1 va X = - 1 la nghiem ciia (1); (d) X = 1 va X = -1782 la nghiem ciia (1). 9.
Cho phuong trinh x+.l=-^ x+1
(1)
( a ) ( l ) « x ^ + 3x + 2 = 0; (b)(l)o(x-l)(x-2); ( c ) ( l ) o ( x - l ) ( x + 2); (d)(l)o(x+l)(x-2). Hay chgn ket qua diing. 10.
Cho phuong trinh '
x + Vx = 0
(1)
( a ) ( l ) o x = O v a x = 1; ( b ) ( l ) c ^ x = 0; (c)(1) < » x = 1 vax = 2; (d)(l)c^x = 0vax = -2. Hay chgn ket qua diing. 11.
Cho phuong trinh
v;^
2 . /- . X + vx +
1
1
x-3
Vx-2 255
(a) Diiu kien ciia phuong trinh la : x > 0, x 9^ 3; (b) Diiu kien ciia phuong trinh la : x > 2; (c) Diiu kien eiia phuong trinh la : x > 2, x ?!: 3; (d) Diiu kien ciia phuong trinh la : x > 0. Hay chgn ket qua diing. 12.
Cho hai phuong trinh : |x|= 1 (1) va x^ - 3x + 2 = 0 (2) (a) (1) la ha qua ciia (2); (b) (2) la he qua cua 1); (c)(l)«(2); (d) ca ba ket luan tren diu sai. Hay chgn ket qua diing.
13.
Cho hai phuong trinh x+l + ^ i = = - 2 Vx + 1
(1)
va x^ + 2x + 5 = 0 Hay chgn ket qua diing trong cac khing dinh dudi day: (a) (1) la he qua cua (2); (b) (2) la he qua cua 1); (c)(l)c^(2); (d) ca ba ket luan tran diu sai. 14.
256
x2 1 . = . cd nghiem la Vx-1 Vx-1 (a) X = 1 va X = - 1 ; (b) X = 1 va X = 0; (c)x=l; (d) X = 1 va X = - 2 . Hay chgn ket qua diing.
Phuong trinh
(2).
15.
Phuang trinh
^
Vx^
Vx^"
(a) cd nghiem x = 2; (b) vd nghiem; (c) cd nghiem x = 3; (d) ca ba ka't luan tran diu sai. Hay chgn ket qua diing. 16.
Cho phuang trinh : x"^ + 3x^ + 2 = 0. (a) Phuang trinh cd 2 nghiem x = - 1 va x = - 2 ; (b) Phuong trinh cd 4 nghiem x = ±1 va x = ±2; (c) Phuang trinh vd nghiem; (d) ca ba ket luan tran diu sai. Hay chgn ka't qua diing.
17.TKBGOAIS610NC-T1
§2. Phufdng trinh bae nhat va bae hai mot an •
'V^i
(tiet 3, 4) L
MUC TIEU
1.
Kien thirc Cling cd kia'n thiic vl va'n dl bien doi tuong duong cac phuong trinh. Nim dugc cac iing dung ciia dinh If Vi-et.
2.
KT nang Nim viing each giai va bien luan phuong trinh bae nha't: ar + b = 0 va bae hai
ax + bx + c = 0. Bia't each bien luan sd giao diem cua mdt dudng thing va mdt parabol va kiem nghiem lai bing do thi. Biet ap dung dinh If Vi-et de xet dau ciia mdt phuong trinh bae hai va bien luan sd nghiem cua mdt phuong trinh triing phuang. 3.
Thai do • Ren luyen tfnh cin than, dc tu duy Idgic va to hgp.
IL
CHUAN BI CUA GV VA HS
1.
Chuan bi cua GV : Chuin bi bai kT cac kien thiie ma HS da hgc d ldp 9 de dat cau hdi. Chuin bi mdt sd hinh ve trong SGK va pha'n mau,...
2.
Chuan bj ciia HS : • Can dn lai mdt so kien thiic vl ham sd da hgc. Can dn lai phin phuong trinh da hgc d ldp 9 va bai 1.
258
in.
PHAN
PHOI THCII LI/ONG
Bai nay 2 tiet : Tie't thic nhdt: tu: ddu de'n he't phdn 3; Tie't thic hai: td phdn 4 de'n he't phdn bdi tap. IV. TIEN TRINH DAY HOC A. Bai cii cau hoi 1 Hay tim nghiem cua cac phuong trinh sau : a) X - 1 = Vx - 1 ; b ) x ' - 3 x + 2=0; c) V 3 - X = 4 x-3 cau hoi 2 Phuong trinh mx^ + x - 1 = 0 ludn cd hai nghiem. Diing hay sai? B. Bai mdi HOATDONGI GV: Ddt van de : Phuang trinh bdc nhdt (mdt an), tiic la phuong trinh cd dang ax+ b = 0 (a va b la hai sd da cho vdi a ^ 0); Phuang trinh bae hai (mot in) la phuong trinh ed dang ax + bx + c = 0 (a, bvacla
ba sd da cho vdi a ^ 0); 2
Ta cd: A = b^ b - 4ac ggi la Me Met thitc. A' = b'^ - ac (vdi b = 2b') ggi la Met thdc thu ggn eiia phuong trinh bae hai.
259
Trong bai nay, chiing ta se nghidn ciiu each giai va bien luan cac phuong trinh cd dang phuong trinh bae nha't hoac phuang trinh bae hai va cd chiia tham sd. GV: Thuc Men thao tdc ndy trong 5 phiu. Hoat ddng cua HS
Hoat ddng cua GV c a u hdi 1 Hay giai phuang trinh : 3x - 1 = X - 3 c a u hdi 2 Cho phuang trinh x^ - 4 X + 1 = 0 Hay giai phuang trinh nay bing 2 each : tfnh A hoac A'
Ggi y tra Idi cau h d i l 2x = - 2 <^ X-= -1 Ggi y tra Idi cau hdi 2 Cach 1 A = 16-- 4 = 12 Xi =
4 + 2V3 = 2 + VI 2
4--2V3 = 2 - VI 2 Cach 2 A' =: 4 - - 1 = 3 X2 =
-VI. X2 =--2- -VI X l - :24
1.
Giai va bien luan phuong trinh dang ax + b = 0
2) a = 0 va b 9^ 0 :
Phuong trinh cd nghidm duy nhat x = — . a Phuong trinh vd nghiem.
3) a = 0 va b = 0 :
Phuong trinh nghiem diing vdi mgi x e R.
l)a^O:
GV: Thuc hien thao tdc ndy trong 3 phut. Hoat ddng ciia GV
cau hdi 1 Cho phuang trinh : (m'- l ) x - m + l =0. 260
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 2
a=m -l;b = -m+l.
Hay xac dinh cac he sd a va b.
cau hdi 2
Ggi y tra Idi cau hdi 2 Hay giai va bien luan theo m Khi a ^ 0 <» m # ±1, phuong trinh cd phuang trinh nay. nghiam duy nhit x = m+ 1 Khi m = 1 : ta thiy a = 0, b = 0 phuong trinh cd vd sd nghiem. Khi m = - 1 : a = 0, b = 2 7^ 0 phuang trinh vd nghiem.
GV : Hudng ddn HS ldm vi du 1. GV hudng ddn HS gidi bang cdch ddt vdn de. HI. Phuong trinh da cho tuong duong vdi phuang trinh nao? H2. Hay chia cac trudng hgp va bien luan. H3. Ka't luan nghiem. HOAT DONG 2 2.
Giai va bien luan phuang trinh dang ax^ + bx + c = 0 1) a = 0 : Trd vl giai va bien luan phuong trinh bx + c = 0. 2) a ?!: 0 : Neu A < 0, phuong trinh vd nghiem; Na'u A = 0, phuong trinh cd mdt nghiem (kep) x = -
2a
Na'u A > 0, phuong trinh cd hai nghiem (phan biet) GV: Thitc Men thao tdc ndy trong 3' Hoat ddng cua GV
cau hdi 1 Cho phuang trinh : x^-2x + m - l = 0 Hay xac dinh cac he so a va b.
Hoat ddng cua HS Ggi y tra Idi cau hoi 1 a = 1; b = -2; c = m - 1.
261
cau hdi 2
Ggi y tra Idi cau hdi 2 Hay giai va bien luan theo m A' = l - m + l = 2 - m . phuang trinh nay. Neu A' < 0 <:5' m > 2 phuong trinh vd nghiem. Na'u m = 2 phuang trinh cd nghiem kep X - 1. Na'u m < 2 phuong trinh cd 2 nghiem phan biet.
GV: Hudng ddn HS ldm vi du 2. GV hudng ddn HS gidi bang cdch ddt van de. HI. Phuong trinh tran da la phuong trinh bae hai chua? H2. Chia cac trudng hgp va bien luan. H3. Tfnh A neu dd la phuong trinh bae hai. H4. Ka't luan nghiem. Chd y 1
•>
,
Trong vf du 2, khi m = 4 ta cd Xj = X2 = — nan ta cd tha chi nau ba trudng hgp : m > 4 ; 0 ; ^ m < 4 v a m = 0 (trudng hgp 0 ^^ m < 4, phuong trinh co nghiem la Xj va X2). GV: Hudng ddn HS thuc hien\\\^ Hoat ddng ciia GV
trong 3 phut Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 c a u hdr 1 Phuang trinh da cho cd the vd Khdng vi ta thiy x = 1 la nghiem. nghiem dugc hay khdng? c a u hdi 2 Ggi y tra Idi cau hdi 2 Phuang trinh ludn cd 2 Tiir phuong trinh nghiem phan biet cd diing X - mx + 2= 0. khdng? tathay : (1 - m ) x = - 2 Neu m = 1, phuong trinh nay vd nghiem, phuong trinh da cho cd | 262
nghiem duy nhat x = 1. Na'u m 7^ 1, phuong trinh nay cd 2 2 va neu =1 nghiam x = m-1 m-1 <:^ m = 3 phuong trinh nay cd nghiem X = 1, phuong trinh da cho cd nghiem kep X = 1. Vdi m 9^ 3, phuang trinh da cho cd hai nghiem phan biet. GV : Hudng ddn HS ldm vi du 3. HI. Hay dua phuong trinh vl dang f(x) = a. H2. Hay khao sat va ve do thi ham sd y = f(x). H3. Bien luan so nghiem phuong trinh bing do thi. GV: Hudng ddn HS thuc Men |H2|. GV treo hoac vehinh 3.1 len bdng vd thitc Men thao tdc ndy trong 3 phiu. Hoat ddng ciia GV
Hoat ddng cua HS
cau hdi 1 Ggi y tra Idi cau hdi 1 Dua vao hinh 3.1, tim cac gia Phuong trinh (3) cd nghiem duong khi tri ciia a de phuang trinh (3) phuong trinh (4) cd nghiem duong. cd nghiem duang. Dua vao hinh 3.1 ta tha'y a > 2 phuong trinh cd nghiem duong. c a u hdi 2 Ggi y tra Idi cau hdi 2 Trong trudng hgp dd, hay tim Nghiem duang cua phuang trinh la nghiem duang ciia phuang nghiem ldn ciia phuang trinh (4). trinh. Giai (4) ta dugc x = -1 +V2 + a . Chu y Khi via't phuong trinh (3) dudi dang x^ + 3x + 2 = x + a, ta thay ka't qua tran con cho bia't sd giao diem cua parabol y = x + 3x + 2 vdi dudng thing y = X + a.
263
HOATDONG3 3.
tTng dung ciia dinh Ii Vi-et
6 ldp dudi, chiing ta da hgc dinh If Vi-et dd'i vdi phuong trinh bae hai va cac ling dung cua nd. Ta nhic lai hai iing dung quan trgng sau : 1) Na'u da thiic f(x) = ax + bx + c cd hai nghiem Xj va X2 thi nd cd th6' phan tfch thanh nhan tii f(x) = a(x-Xi)(x-X2). GV: Thuc Men thao tdc ndy trong 3 phiit. Hoat ddng ciia GV c a u hoi 1 Tim nghiem ciia da thiic :
Hoat ddng cua HS Ggi y tra Idi cau hdi 1 Da thiic da cho cd nghiem la : x = 1, X = 6
f(x) = x' - 7x + 6 Cau hdi 2 Hay phan tfch da thiic thanh Ggi y tra Idi cau hdi 2 nhan tix. f(x) = ( x - l ) ( x - 6 ) .
2) Ne'u hai so cd tdng la S va tfch la P thi chiing la cac nghiem ciia phuang trinh x^ - Sx + P = 0. GV: Hudng ddn HS thuc hien\HZ\ vd thuc Men thao tdc ndy trong 5 phiit. Hoat ddng ciia GV
cau hdi 1
Hoat ddng cua HS
Ggi y tra Idi cau hdi 1 Ne'u ggi chilu dai va chieu a + b = 20 rdng ciia hinh chir nhat la a va a. b = P b thi ta cd bieu thiic nao ? Ggi y tra Idi cau hdi 2 c a u hdi 2 Hay lap phuang trinh cd hai a) Vdi P = 99, phuang trinh la x^ nghiem la a va b trong timg 20x + 99 = 0,Xi=9,X2=ll. trudng hgp. Ta phai khoanh hinh chu nhat kfch thudc 9 cmxll cm. 264
b) Vdi P = 100, phuong trinh la x ^ - 2 0 x + 1 0 0 = 0, Xi=9, X2= 11. Ta phai khoanh hinh chii nhat kfch thudc 10 cmxlO cm. c) Vdi P =101, phuong trinh la X - 20x + 101 = 0, vd nghiem. Vay trong trudng hgp nay, khdng cd hinh chir nhat nao thoa man yau ciu di bai. Nhan xet dudi day suy ra true tiep tut dinh If Vi-et. Nd cho phep ta nhan bia't da'u cac nghiem ciia mdt phuong trinh bae hai ma khdng cin tfnh cac nghiem dd. 2
Cho phuang trinh bae hai ax + bx + c = 0 cd hai nghiem Xj va X2 vdi gia b c thiat rang Xj < X2. Kf hiau S = — va P = —. Khi dd : a
a
- Na'u P < 0 thi Xj < 0 < X2 (hai nghiem trai da'u). - Na'u P > 0 va S > 0 thi 0 < x^ < X2 (hai nghiem duong). - Na'u P > 0 va S < 0 thi Xj < X2 < 0 (hai nghiem am). Vi du 4. Phuang trinh ( l - V 2 ) x ^ - 2 ( l + V2)x + V 2 = 0 cd : a = l - V 2 < 0 va c = V 2 > 0 => P<0. GV: Huang ddn HS van dung chii y de gidi cdc vi du 5 vd 6. Chu y Trong vf du tran, ca hai ka't luan (phuong trinh cd hai nghiem, va hai nghiem do trai da'u) diu dugc suy ra tii P < 0.
265
Trudng hgp P > 0, ta phai tfnh A (hay A') de xem phuong trinh cd nghiem hay khdng roi mdi tfnh S di xac dinh da'u cac nghiem. GV: Hudng ddn gidi vi du 5. Sau dd, GV hudng ddn HS thuc hien\H4\ (GV thuc hien thao tdc ndy trong 5 phiit). Hoat ddng cua HS
Hoat ddng ciia GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 Hay xet da'u ciia a va c trdng a va c trai dau phuang trinh a) Ggi y tra Idi cau hdi 2 c a u hdi 2 phuang trinh a) cd hai nghiem Cd ka't luan gi ve nghiem trai da'u. phuang trinh a). Chgn (A) Cau hdi 3 Ggi y tra Idi cau hdi 3 Hay lam tuang tu dd'i vdi b) Phuong trinh phuang trinh b) x^ - ( V 2 + V I ) x + Vd = 0 cd hai nghiem phan biet Xj va X2 vi: A =(V2 + ^ ) ^
-
4V6=
(2
-
2V6+3)-4V6= (V2-VI) >0 Do
X1X2 =—= V 6 > 0
a
va Xi+ X2
= — = V 2 + V I > 0 nan hai nghiam nay a cung da'u duong. Do do ta chgn phuong an (B). Ta da bia't: ddi vdi phuong trinh triing phuong ax^ + bx^ + c = 0,
(4)
neu dat y = x (y > 0) thi ta di den phuong trinh bae hai dd'i vdi y : ay^ + by + c = 0. 266
(5)
Do dd, mudn biet sd nghiem cua phuong trinh (4), ta chi cin biet sd nghiem cua phuong trinh (5) va diu cua chiing. GV: Hudng ddn HS thitc Men H 5 va thuc Men thao tdc ndy trong 5 phiu. Hoat ddng cua HS Hoat ddng ciia GV Ggi y tra Idi cau hdi 1 c a u hdi 1 a) Na'u phuang trinh (4) cd Diing nghiem thi phuang trinh (5) chic chin cd nghiem. Dung hay sai? Ggi y tra Idi cau hdi 2 cau hdi 2 b) Sai, vi khi phuang trinh (5) chi cd b) Na'u phuang trinh (5) cd nghiem am (hoac mdt nghiem kep am, nghiem thi phuang trinh (4) hoac hai nghiem am phan biet) thi chic chin cd nghiem. phuong trinh (4) vd nghiem. Diing hay sai? GV: Hudng ddn HS ldm vi du 6 trong SGK. Ngodi ra GV cd the cho HS ldm them vi du sau : Vi du. Cho phuang trinh : x"^ - 2(m + 3)x^ + 2m + 5 = 0.
(6)
Vdi gia tri nao cua m thi phuong trinh (6) cd bdn nghiem phan biet? Gidi: 2
Dat y = X (y > 0), ta di dan phuong trinh : y^ - 2(m + 3)y + 2m + 5 = 0.
(7)
Hien nhian, phuong trinh (6) cd bd'n nghiem phan biet khi va chi khi phuang trinh (7) cd hai nghiem duong phan biet. Vi (7) cd hai nghiem y = 1 (do a + b + c = 0) va y = 2m + 5 nan diiu kien de phuong trinh (6) cd bd'n nghiem phan biet la 0 < 2m + 5 ;>t 1 hay - 2,5 < m ^ - 2 . 26/
TOM T A T B A I HOC 1.
Giai va bien luan phuong trinh dang ax + b = 0 1) a 9t 0 :
2.
2) a = 0 va b ;^ 0 :
Phuong trinh cd nghiam duy nhat x = — • a Phuang trinh vd nghiem.
3) a = 0 va b = 0 :
Phuong trinh nghiem diing vdi mgi x e R.
Giai va bien luan phuong trinh dang ax^ + bx + c = 0 1) a = 0 : Trd vl giai va bien luan phuong trinh bx + c = 0. 2) a ?^ 0 : Na'u A < 0, phuong trinh vd nghiem; Na'u A = 0, phuong trinh cd mdt nghidm (kep) x =
; 2a
Neu A > 0, phuang trinh cd hai nghiem (phan biet) 3.
VI dinh If Vi-et. ' 1) Na'u da thiic f(x) = ax + bx + c cd hai nghiem Xj va X2 thi nd cd the phan tfch thanh nhan tu f(x) = a(x - Xi)(x - X2) (xem bai tap 9). 2
2) Na'u hai sd cd tong la S va tfch la P thi chiing la cac nghiem ciia phuang trinh x - Sx + P = 0. 2
3) Cho phuong trinh bae hai ax + bx + c = 0 cd hai nghiem Xj va X2 vdi , . b e gia thiet rang Xj < X2. Kf hiau S = — va P =— Khi dd : a
a
- Ne'u P < 0 thi Xj < 0 < X2 (hai nghiem trai da'u). - Neu P > 0 va S > 0 thi 0 < Xj < X2 (hai nghiem duong). - Neu P > 0 va S < 0 thi Xj < X2 < 0 (hai nghiem am).
268
HOATDONG 4 HUONG DAN
GIAI BAI
TAP SGK
Bai 5. GV: Hudng ddn gidi cdu a) Hoat ddng cua GV
Hoat ddng ciia HS
c a u hdi 1
Ggi y tra Idi cau hdi 1
Cach giai a) diing chua? c a u hdi 2
Sai vi chua cd tap xac dinh. Ggi y tra Idi cau hoi 2
Ka't luan nghiem diing hay Sai, vi chua so sanh nghiem vdi tap tap sai? xac dinh ciia phuang trinh. Trd Idi b) h) Sai vi khi binh phuong hai va' ta chi dugc phuong trinh he qua. Cin thuf lai nghiem (gia tri tim dugc cua x khdng thoa man).
Bai 6. GV: Hudng ddn gidi cdu a) Hoat ddng ciia GV c a u hdi 1 phuang trinh nay da phuang trinh bae hai chua
Hoat ddng ciia HS Ggi y tra Idi cau hdi 1 la Chua vi ha sd a chiia tham sd.
c a u hdi 2 Ggi y tra Idi cau hdi 2 Hay giai va bien luan phuang 2m-3 . X =— (voi mot m) trinh. m^+1 Trd Idi cdc cdu edn Iqi b) Vdi m = 1, phuong trinh nghiem diing vdi mgi x
269
Vdi m 9t 1, phuong trinh cd nghiem duy nhat x = m + 2. c) Vdi m ^ 2 va m ;t 3, phuong trinh vd nghiem Vdi m = 2 hoac m = 3, phuong trinh nghiem diing vdi mgi x. Ggi y m(x - m + 3) = m(x - 2) + 6 <:5' Ox = m^ - 5x + 6 O Ox = (m - 2)(m - 3) d) Vdi m ^ 1 va m :?!: 2, phuong trinh cd nghiem x = m m-2 Vdi m = 1, phuong trinh nghieni diing vdi mgi x. Vdi m = 2, phuong trinh vd nghiem. Ggi y : m (x - 1) + m = x(3m - 2) <» (m - 3m + 2)x = m'
m
c^ (m - l)(m - 2)x = m(m - 1) Bai 7. GV:
Hudng ddn giai cdu a) Hoat ddng cua GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Phuang trinh nay da la Chua vi ha sd a chiia tham so. phuang trinh bae hai chua? c a u hdi 2 Goi y tra Idi cau hdi 2 Trong nhiing trudng hgp nao Phuong trinh cd mdt nghidm trong 2 thi phuang trinh dang ax + mdi trudng hop sau : bx + c = 0 cd mdt nghiem? a = 0 va b 9^ 0 a ;t 0 va A = b^ - 4ac = 0 Trd Idi cdc cdu con Iqi b) Phuong trinh vd nghiem trong m 5i trudng hgp sau : a = b = Ovac;tO a ;^ 0 va A = b^ - 4ac < 0.
270
Bai 8. GV:
Hudng ddn giai cdu a) Hoat ddng cua GV
cau hdi 1 Phuang trinh nay da phuang trinh bae hai chua?
Hoat ddng cua HS Ggi y tra Idi cau hdi 1 la Chua vi he sd a chira tham sd.
c a u hdi 2 Ggi y tra Idi cau hdi 2 Giai va bien luan phuang Khi m = 1, phucmg trinh trd thanh trinh a) 3x - 1 = 0, cd mdt nghiem x = — Khi m :; — - phuong trinh cd hai 4 -3±V4m + 5 nghiem Xl 2 = ; 2fm-lj Vdi m < — - phuong trinh vd nghiam. 4 Trd Idi cdc cdu edn Iqi b) DS : X = 2 ± V7 - m neu m <7. vd nghiem na'u m > 7.
Bai 9. GV: Hudng ddn gidi cdu a) Hoat ddng ciia GV
cau
hdil
Chirng minh ring
Hoat dong ciia HS Ggi y tra Idi cau hdi 1
phuong Thay vao ta thay ngay Xj va trinh a(x - x,)(x - X2) = 0 ed la cac nghiem cua phuong trinh.
X2
271
hai nghiem Xj va X2. c a u hoi 2 Chiing td :
Ggi y tra Idi cau hoi 2 Vi hai phuong trinh 2 ax + bx + c = 0 va
ax^ + bx + c = a(x - Xi)(x - X2).
a(x - Xi)(x - X2) = 0 tuong duong. Trd Idi cdc cdu con lai Cdch khdc cdu a) h c a) Ta cd cac ha thiic Xi + X2 = — va X1X2 =— Do dd : a a / .2 b \ ax + bx + c = a X + — x + - = a[x - (xi + X2)x + X1X2] V a a; = a(x-Xi)(x-X2) Cdu b) . 1 b) • f(x) cd hai nghiem la - 4 va — nan cd phan tfch f(x) = -2(x + 4)(x - 1/2) = (x + 4)(1 -2x)
V2
• g(x) cd hai nghiem la V2 va —j=— nan cd phan tfch
V2 + I
g(x)=(V2 + l)(x-V2)
X-
41 V2+1,
= (x-V2)r(V2+l)x-V2 Hien nhian phuong trinh cd hai nghiem. Ta cd Xi+ X2 = 2 va X1X2 =-15.
272
Bai 10. Hudng ddn Hien nhian phuang trinh cd hai nghiem. Ta cd XY+ X2 = 2 va XjX2 = -15. a) x\+x\=
(xi+ X2)^ - 2x1X2 = 4 +.30 = 34.
b) Xl +x^= (xi+ X2)^ - 3 X1X2 (xi+ X2) = 8 + 90 = 98. c) x\+x\=
[x\+xl^
-2(XiX2)^ = 34^-2(-15)^ = 706.
Bai 11. Chgn (B). Hudng ddn : Chii y ring phuang trinh bae hai tuong iing cd ac < 0 nan nd cd hai nghiem trai da'u, suy ra phuong trinh da cho cd diing hai nghiem dd'i nhau. Tii do ta loai cac phuang an (A) va (C). Phuang an (D) ciing bi loai bing each thii true tia'p.
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273
BO SUNG KIEN THtTC De giai phuong trinh, ddi khi ta diing each dat in phu. Gia sii ta cin giai phuang trinh sau vdi diiu kien D : F(x) = 0
(1)
Neu F(x) cd the viet dugc dudi dang F(x) = f(u), trong dd u = g(x) thi viec giai phuong trinh (1) cd the tien hanh nhu sau : Budc 1 : Dat in phu. Trong phuang trinh (1), dat u = g(x) ta cd phuang trinh an la u (thudng ggi la phuong trinh trung gian) : f(u) = 0
(2)
Budc 2 : Giai phuong trinh (2). - Neu (2) vd nghiem thi (1) vd nghiem. - Neu (2) cd cac nghiem la Ui, U2,... thi vdi diiu kien D : gfxj = ui (1) O
gfxj = U2
Budc 3 Giai cac phuong trinh g(x) = Ui, g(x) = U2,... vdi diiu kien D. Khi dd tap nghiem ciia phuong trinh (1) la tap tat ca cac nghiem thu dugc. Chu y : Hai phuang trinh (1) va (2) la nhiing phuong trinh khdng ciing in nan ta khdng the viet (1) <^ (2). Vi du : Giai phuong trinh (x^ + x)^ - (x^ + X) - 12 = 0
(3)
Gidi : Trong phuong trinh (3), dat u = x + x ta cd phuong trinh an u : u^ - u - 12 = 0 274
(4)
Giai phuang trinh (4) ta dugc hai nghiem la Ui = - 3 va U2 = ^. Dodo (3) <» X + X = - 3 hoac x + x = 4 Ta cd: 2
2
X + x = -3<=>x + x + 3 = 0 (phuong trinh nay vd nghiem) 2 . 2 . n -1±V17 X + x = 4 <» x + x - 4 = 0 <:^ x = - 2 Ka't luan : Phuang trinh (3) cd hai nghiem x =
-i±Vn
275
MOT SO BAI TAP TRAC NGHIEM 1.
Cho phuong trinh m(x - 2) = 3x + 1 (a) Phuang trinh ludn cd nghiem; (b) Phuang trinh ludn cd nghiem
v6im^3; m -I- 1
(c) Phuang trinh ludn cd nghiam x =
m-3
;
(d) Phuang trinh ludn cd hai nghiem. Hay chgn ket qua diing. 2
2.
Cho phuang trinh mx + 2mx + m = 0. (a) Phuang trinh ludn cd hai nghiem phan biet Vm; (b) Phuang trinh ludn cd nghiem kep Vm ^ 0;
3.
(c) Phuang trinh vd nghiem Vm; (d) ca ba ka't luan tran diu sai. Cho phuang trinh bae 2 : x^ + 3x + 2006 = 0 (a) Phuang trinh cd 2 nghiem Xi va X2 thoa man : Xi + X2 = -3 va Xi.X2 = 2006; (b) Phuang trinh cd 2 nghiem Xi va X2 thoa man : Xj + X2 = 3 va X1.X2 = 2006; 3 (c) Phuang trinh cd 2 nghiem Xj va X2 thoa man : Xi + X2 = — va X1.X2 = 2006; (d) Phuang trinh vd nghiem. Hay chgn ket qua diing.
4.
Cho hai so x va y thoa man bd phuong trinh : rx + y = 2 lx.y = l
276
Khi dd : (a)
'x = l
(b)
[x = - l [y = - i
x=l y=l
(c)
'x = 0 y = 2'
'x = 2 y=0
(d) khdng tdn tai x va y. Hay chgn ket qua diing. Cho 2 sd X va y thoa man : X + y = 3 va x.y = 2 Khi dd : (a)
(b)
(c)
'x = - 2 y=l
x=l y = -2
^x = l .y = i 'x = 2 y= l'
x=l y=2
(d) khdng ton tai x va y. Hay chgn ket qua diing. 6.
Cho 2 sd X va y thoa man ha phuang trinh :
x^+y^=5 x+y=-1
Khidd (a)
(b)
ly = 2' x= -l y=2
x=2 y=l 'x = 2
.y--i 277
(c)p^\; 1^ = -^ ly = -2 fx = l [y = -2 7.
[y = l
Cho phuong trinh bae hai ax + bx + e = 0 ed 2 nghiem Xi va X2 khi do phuong trinh bae hai cd 2 nghiam la — va — la M
X2
(a) -ex - bx - a = 0 (b) CX + bx + a = 0 (c) CX - bx - a = 0 (d) cx^ - bx + a = 0. 2 8.
Cho phuang trinh bae 2 : x |xi -X2I bing
+ 3 x - 1 2 = 0 c d 2 nghiem Xi va X2. Khi do
(a)V5^
(b)57
(c)-V57
(d)V57 2
9.
10.
Phuang trinh x - 2mx + m + l = 0 c d 2 nghiem X|, X2 ma Xi - 3x2 ~ ^ khi dd m bang (a)m = 2
(b)m=--
2 (c) m = 2 hoac m = —
(d) cac ka't qua tran dau sai.
Phuang trinh |x - 1 | + x = 2 3 (a) cd nghiem la x =— ; (c) cd nghiem la x = — hoac x = 1;
11. 278
(b) vd nghiem; (d) ca ba ket luan tran diu sai.
Phuang trinh |x - ij + |x +1| = 2 cd nghiem la
12.
(a)-lhoacl;
(b) x = 1 hoac x = - ;
(c)x = - l h o a e x = - ; 2 Hay chgn ket qua diing.
(d)-1 < x < 1.
Phuong trinh x + 1 = 4x + 2 ed nghiem la (b)K=^i^hoacx=---i±^ 2 2
(a)x=0;
,.
(c)x=
- 1 + V5
...
;
-l-VI
(d)x=-—
Hay chgn ket qua diing. 13.
14.
Cho phuong trinh x + Vx - 1 = 1 cd nghiem la (a) X = 0 hoac x = 1;
(b) x = 1;
(c) X = 0 hoac X = - 1 ;
(d) x e 0 .
Cho phuong trinh x + 2x = Vx + 1 - 1 (1) cd nghiem la (a)x = - l ;
(b)x = 0;
(c) X = 2;
(d) X = - 2 .
Ddp dn: l.(b).
2. (b).
3. (d).
4. (a).
5. (c).
6. (c).
7 (d).
8. (d)
9. (c).
10. (a).
11. (d).
12. (b).
13. (b).
14. (a).
BAITAPTVGIAI Hay chgn ket qua dung trong cdc ke't qua sau ddy: 15.
16.
Phuong trinh (m - l)x + x - 2 = -x + 1 cd nghiem duy nhit khi (a)m ^ 1;
(b) m
(c) m ?t 2;
(d) m ;i - 2 .
Phuang trinh 2x + 1 = 3x + 2 cd nghiem la (a)x=l;
(b)x = 2;
(c)x = - l ;
(d)x = - 2 . X
17.
^-\;
1
Phuang trinh
= 1 cd nghiam la x+1
•
2x-l
(a) X = 0 hoac x = 1;
(b) x = 0 hoac x = - 1 ;
(c) x = - 1 hoac X = 1;
(d) x = 0.
2
18.
Cho phuong trinh x + 2mx + m + 1 = 0 cd nghiem kep khi 1-V5 , . - 1 + V5 hoac m = ; (a) m = 2 ' 2
(u^
-1-V5 , .
(b) m =
- l + VI
hoac m = 2
2
, , 1-V5 (c) m =
, . 1 + V5 hoac m = ;
2
2
(d) m =
hoac m = 0. 2 2
19.
Phuong trinh x - 3 x + l = 0 c d 2 nghiem Xi va X2 thoa man . . fxi+X2 =3
(a)
'
'
(b) M
2
[xi.X2=l
[xi.X2=2
f x i +X2 = 3
f x i +X2 = 3
(c) ^ ^ • [xi.X2=-l 280
fxi +XT =3
^^A
\ [xi.X2=0.
20.
21.
Phuong trinh x^ - 3x + V5 = 0 cd 2 nghiem |xi - X21 bing (a)V9 + 4V5
(b)V9-4V5;
(c)9 + 4V5
(d)9-4V5
Phuang trinh x^ + 2 V 5 x + V s =Ocd 2 nghiem Xl vax2, Xl + X2 bing (a)20-V5;
(b)V20-V5 ;
(c)20+V5; (d)V20 + V5 ~ 2 2 2 Hudng ddn : Xj + X2 = (xj + X2) - 2xi.X2 22.
Phuang trinh 2x - 3x - 1 = 0 cd 2 nghiem Xi va X2 ma Xi + X2 bing (a)-
(b)-;
9 (0-
12 (d)-
23.
3
3
(a)x=^ (c) X = 1 va X = - 2 24.
25.
2
Huang ddn : Xl + X2 = (xi + X2)[(xi + X2) -3x1X2]. Phuong trinh 2x + 1 + V2x = 3 cd nghiem la (b)x = 2; (d) X = 0.
Phuong trinh |x + l | + 2 x + l = 0 c d nghiem la (a)x=-
(b)x = 0;
(c)x=l
^"^^""""f
Phuong trinh x^ = -Vx - 1 (a) cd nghiem X = 0;
(b) cd nghiem x = 1;
(c) vd nghiem;
(d) cd nghiem x = - ! .
281
26.
27.
28.
Phuong trinh | x - 2 | + | x - 3 | = 5 c d nghiem la (a) X = 0
(b) X = 5;
(c) 0 < X < 5
(d) 0 < X < 5.
Phuong trinh x^ - 3x^ - 4 = 0 (a) cd 2 nghiem;
(b) cd 4 nghiem;
(c) cd 3 nghiem;
(d) vd nghiem.
Phuong trinh x + mx + m - l = 0 c d 4 nghiem each diu nhau khi (a)m =
-5-V35
(b)m=-
-5 + V35
(d)m = 7+V35
(c) m = 5 +V35 ;
Hudng ddn .• Dat X^ = X ^ X^ + mX + m - 1 = 0 (2) =^ (2) cd 2 nghiem Xi va X2 sao cho X2 = 9Xi. 4
29.
Phuong trinh x - 3x + 2 = 0 cd cac nghiem la (a)lva2; ( b ) - 2 ; - 1 ; 1 va 2; (c)-lva-2;
282
2
( d ) - 2 va 2.
Luyen tap (tielt 5, 6) I.
MUC T I E U
1.
Kien thiifc Giiip HS
Cling cd cac kien thiic da hgc trong bai 2 vl phuong trinh bae nhat va phuong trinh bae hai. 2.
KI nang
Ren luyen cac ki nang : giai va bien luan phuong trinh bae nhat hay bae hai mdt in cd chira tham so; bien luan sd giao diem cua dudng thang va parabol; cac iing dung ciia dinh If Vi-et, nhat la trong viec xet dau cac nghiem cua phuong trinh bae hai va bien luan sd nghiem ciia phuong trinh trung phuong. On tap each giai phuong trinh bae nhat. Giai thanh thao phuong trinh bae hai va tham phuong phap nhim nghiem. 3.
Thai do • Tao cho HS tfnh cin than, dc tu duy Idgic va tdng hop tdt. Lidn ha dugc vdi thuc te.
II.
CHUAN BI CUA GV vA HS
1.
Chuan hi ciia GV : Chuin bi bai kT cac bai tap (trgng tam la cac bai 12, 16, 17, 18, 19, 20). Chuan bi mdt sd hinh ve trong SGK, phan mau,...
2.
Chuan bi cua HS : Can dn lai mdt sd kien thiic vl bai 1, xem lai cac [llj, cac vi du va lam
bai trudc tai nha.
283
m.
PHAN
PHOI THCII LUONG
Bai nay 2 tia't : Tie't thd nhdt: Chua vd hudng ddn de'n bdi 18; Tiet thic hai: phdn con lai. IV.
TIEN TRINH DAY HOC
A. Bai cii cau hoi 1 Hay tim nghiem cua cac phuong trinh sau : a)x-4=Vx^;
b) x ^ - 3 x - m + 2 = 0;
c ) V I ^ = Vx2-4x-3 cau hoi 2 Phuong trinh x^ + x + m = 0 ludn vd nghiem. Diing hay sai? B. Bai mdi HOATDONGI Bai 12 a). Hoat dong ciia GV c a u hoi 1 Bia'n doi phuang dang bae nha't.
trinh
c a u hdi 2 Bien luan phuang trinh tren.
Hoat dong cua HS Ggi y tra Idi qau hdi 1 vl (m + 2)x - 3m - 3 = 0
Ggi y tra Idi cau hdi 2 m + 3 ., ^ . nghiam X= neu m J^ -2, vo m+2 ne'u m = - 2 .
Trd Idi cdu con Iqi: b) X =
neu m ?=: 1, nghiem diing vdi moi x neu m = 1. 3
284
. 5m + l . 1 nau m ^ — - vd nghiem neu m = 3 c) X = 3m+ 1 3 d)x=na'u m Tt ± 2, m+2 vd nghiem neu m = - 2 , nghiem diing vdi mgi x neu m .= 2. HOATDONG 2 Bai 13. GV: Chaa cdu a) Hoat dong cua HS
Hoat ddng ciia GV c a u hdi 1 Bia'n doi phuang trinh ve dang bae nha't. cau hdi 2 Bien luan phuang trinh tren.
Ggi y tra Idi cau hdi 1 p x - 2 = 0. Ggi y tra Idi cau hdi 2 p = 0 phuang trinh vd nghiem.
Trd Idi cdu con Iqi: b) p = 2.
HOAT DONG 3 Bai 14. GV: Chaa cdu b) Hoat ddng cua GV
Hoat ddng cua HS
c a u hoi 1 Ggi y tra Idi cau hdi 1 Tun nghiem diing ciia phuong -2VI-V7 -2VI + V7 trinh
42x^ + A43X-242
=0.
'^~
V^
''^
4i 285
Cau hdi 2 Tim cac nghiem gin diing.
Ggi y tra Idi cau hdi 2 j c * 0,38 ; x * - 5 , 2 8 .
Trd Idi cdu con Iqi: a ) x « 4 , 0 0 ; x * 1,60. HOATDONG 4 Bai 15. Hoat ddng cua HS
Hoat dong ciia GV c a u hoi 1 Ggi canh ngin nhat la x. hay lap phuang trinh theo x.
c a u hdi 2 Tim cac canh.
Ggi y tra Idi cau hdi 1 (x + 25)^ = (x + 23)^ + x^ <^ x ^ - 4 x - 9 6 = 0 Ggi y tra Idi cau hdi 2 Riuong trinh nay cd hai nghiem Xi = 12 va X2 = - 8 . Do diiu kien x > 0 nan X2 hi loai. Cac canh la : 12, 35, 37.
HOATDONG 5 Bai 16 b). GV: Thuc Men thao tdc ndy trong 7 ph 'a. Hoat ddng ciia GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Phuang trinh trdn da la Chua, VI ha sd a phu thudc tham sd m. phuang trinh bae hai chua? Ggi y tra Idi cau hdi 2 c a u hdi 2 Khi m = 0 phuang trinh la - 6 x + l = 0 < = > x = — 6 phuang trinh nao? 286
Ggi y tra Idi cau hdi 3 c a u hdi 3 Hay bien luan phuang trinh Vdi m = 1, phuong trinh cd mdt khi m ^ 0. nghiem x = — ; 7
1
Vdi
< m ^ 1, phuong trinh cd
48 - 7 ± V l + 48m (hai 2('m-lj 1 nghiem triing nhau khi m = ); nghiem x
B
•
48
B
Vdi m <
. phuong trinh vd 48
nghiem. Trd Idi cdu con Iqi: 12 a) Vdi m = 1, phuong trinh cd mdt nghiem x = — ; ^r'• Voi
1 ^
, u ^ u ' u-~ - 7 + Vl + 48m 1, phuong trinh eo nghiem x, 2 = ' 2('m-lj (hai nghiem triing nhau khi m =
Vdi m <
1
1
);
phuong trinh vd nghidm.
48 c) Vdi k ?t - 1 , phuang trinh cd hai nghiem x = 1 va x
1 k+1
(khi k = 0, hai nghiem nay trung nhau); Vdi k = - 1 , phuong trinh ed mdt nghiem duy nhat x = 1. d) Khai trien ve trai ta cd phuong trinh m(2m - l)x ^ -(3m - 2)x - 2 = 0 (*); Vdi m = 0, (*) cd mot nghiem x = 1; 287
Vdi m = - . phuong trinh cd mdt nghiem x = 4; 1 2 Vdi m ?!: 0 va m ;^- . phuong trinh cd hai nghiam x = — - x = 2 m
1 2m-l
(hai nghiem nay trung nhau na'u m = - ) .
HOATDONG 6 Bai 17. Hoat ddng cua GV
Hoat dong cua HS
c a u hoi 1 Ggi y tra Idi cau hdi 1 Hoanh do giao diem la nghiem 2x + 2 x - m - 3 = 0 ciia phuong trinh nao? Cau hdi 2 Bien luan sd giao diem.
Ggi y tra Idi cau hdi 2 Day la phuong trinh bae hai vdi biet thiic thu ggn A' = 2m + 7. Dodd: - Khi m < -3,5, (3) vd nghiem, suy ra hai parabol khdng cd diem chung vdi true hoanh. - Khi m = -3,5, (3) cd mdt nghiem (kep), suy ra hai parabol cd mdt diem chung vdi true hoanh. - Khi m > -3,5, (3) cd hai nghiem phan biet, suy ra hai parabol cd hai diem chung vdi true hoanh. HOATDONG 7
Bai 18. Hoat ddng ciia GV
Hoat ddng cua HS
c a u hoi 1 Ggi y tra Idi cau hdi 1 Hay tim diiu kien de phuang Diiu kien d l phuang trinh cd 288
nghiem la A= 4 - ( m - l ) = 5 - m > 0 ,
trinh cd nghiem.
hay m < 5 c a u hdi 2 Ggi y tra Idi cau hdi 2 Khi phuang trinh cd 2 nghiem Xi+ X2 = 4 va X1X2 = m - 1. hay tim tong va tfch hai nghiem dd. Ggi y tra Idi cau hdi 3
c a u hdi 3 Hay bieu diin Xj +X2 theo m.
Xl +X2 = (Xi+ X2)^ - 3 X1X2 (X,+ X2)
= 4^ - 12(m - 1) = 76 - 12m. c a u hdi 4
Ggi y tra Idi cau hoi 4 xl+xl
Tim m.
= 4 0 ^ 7 6 - 1 2 m = 40
<^ 12m = 36 <» m = 3 HOATDONG 8
Bai 19. Hoat dong cua HS
Hoat dong cua GV cau hdi 1 Hay tinh (xj - X2)^
Ggi y tra Idi cau hdi 1 (Xl - X2)^ = 289 => (xi + X2)^ - 4 X1X2 = 289
c a u hdi 2 Tim m.
Ggi y tra Idi cau hdi 2 16m^ + 33 = 2 8 9 ^ m = ± 4 Tii dd ta cd nghiem phuong trinh.
Hudng ddn hoac ddp dn cho cdc bdi tap con Iqi: Bai 20. a) Vd nghiem (vi phuong trinh bae hai cd hai nghiem am); b) Hai nghiem dd'i nhau (vi phuong trinh bae hai cd hai nghiem trai diu) c) 4 nghiem (vi phuong trinh bae hai cd hai nghiem duong phan biet). 19-TKBGBAIS610NC-T1
289
Ggi y : Chii y ring 1 - V2 < 0. d) 3 nghiem (vi phuong trinh bae hai cd mdt nghiem duong va mdt nghiem 0). Bai 21. a) Vdi k = 0, phuang trinh da cho cd mdt nghiem x = -0,5, khdng thoa man diiu kien cua bai toan. Vdi k ;i 0, phuong trinh da cho la phuang trinh bae hai vdi biet thirc A' = k + 1. Do dd nd vd nghiem khi k < - 1; cd nghiem duy nhit x = 0 khi k = - 1 . Ca hai trudng hgp nay diu khdng thoa man dl bai. Cudi cung ta chi edn phai xet hai trudng hgp phuang trinh cd hai nghiem phan biet Xi va X2: k+1 1) Vdi - 1 < k < 0, ta cd X1X2 =
< 0, phuong trinh cd hai nghiam trai
k da'u, nghia la cd mdt nghiem duong (thoa man). k+1 2) Vdi k > 0, ta cd Xi + X2 = X1X2 =
> 0, phuong trinh cd hai k
nghiem duong (thoa man). Ka't luan : Cac gia tri cua k thoa man dl bai la - 1 < k < 0 va k > 0. b) Dat X = y + 1, ta cd phuong trinh kx^-2y-l=0
(1)
Bai toan trd thanh : Tim cac gia tri ciia k de phuong trinh (1) cd hai nghiem trai da'u. Hien nhian, diiu kien de phuang trinh (1) cd hai nghiem trai da'u la k > 0.
290
§3. Mot so phTfdng trinh quy ve phii:ofng trinh bae nhat hoac bae hai (tiet 7) I.
MUC TIEU
1.
Kien thufc Giiip HS :
Nim dugc nhiing phuong phap chii yeu giai va bien luan cac dang phuang trinh nau trong bai hgc. Cung cd va nang cao ki nang giai va bien luan phuong trinh cd chiia tham sd va cd the quy vl phuong trinh bae nhat hoac bae hai. 2.
KT nang On tap each giai phuong trinh bae nhat, bae hai. • Giai thanh thao phuong trinh bae hai va thdm phuong phap nham nghiem.
3.
Thai do Phat trien tu duy trong qua trinh giai va bien luan phuong trinh.
II.
CHUAN BI CUA GV vA HS
1.
Chuan hi ciia GV : • Chuin bi bai ki cac bai tap Chuan bi mdt so hinh ve trong SGK, phan mau,...
2.
Chuan bi ciia HS : Cin dn lai mdt so kien thiic vl bai 1, xem lai cac [Hj va cac vf du. Doc va lam bai trudc d nha.
291
m.
PHAN
PHOI THCil LUONG
Bai nay 1 tiet: IV. TIEN TRINH DAY HOC A. Bai cu c a u hoi 1 Nau cac dinh If vl bien doi phuong trinh tuong duong va phuong trinh he qua. cau hdi 2 Nau cac budc bien lUan phuong trinh dang ax^ + bx + c = 0.
B. Bai mdi HOATDONGI 1.
Phuang trinh dang | ax + b | = | ex + d | Cdch gidi 1 Chiing ta da biet : | X | = | Y | <»X = ± Y (vdi X va Y la sd tuy y). Tuang tu, ta cd : I ax + b I = I ex + d I <:^ ax + b = ± (ex + d).
Nhu vay, mudn giai phuang trinh | ax + b | = | ex + d |, ta chi viec giai hai phuang trinh ax + b = ex + d va ax + b = -(ex + d) rdi lay ta't ca cac nghiem thu dugc. GV: Thuc Men thao tdc ndy trong 2 phut. Giai phuang trinh : jx - 3 1 = |2x +ll Hoat ddng ciia GV
Hoat ddng ciia HS
Ggi y tra Idi cau hoi 1 c a u hdi 1 Phuang trinh tran tuong duong X - 3 = 2x + 1 hoac x - 3 = - 2 x - 1 vdi hai phuong trinh nao? 292
Cau hdi 2 Hay giai phuang trinh trdn.
Ggi y tra Idi cau hdi 2 X - 3 = 2x + 1 <^ X = -4; x - 3 = -2x - 1 2 <=>3x = 2 c : > x = — . 3
GV: Hudng ddn HS ldm vi du 1 GV: Huang ddn HS thuc hien\\\^\ vd thitc Men thao tdc ndy trong 4 phiu. Hoat ddng cua GV
cau hdi 1 Khi m = 1, nghiem phuang trinh la :
Hoat ddng cua HS Ggi y tra Idi cau hdi 1 cua 1 X = —
2
Cau hdi 2 Ggi y tra Idi cau hdi 2 Khi m = - 1 phuang trinh cd 1 nghiem la 2 c a u hdi 3 Khi m 7^ ±1, phuang trinh cd nghiem la :
Ggi y tra Idi cau hoi 3 -m + 2 , m + 2 va m+1 m-1 Hai nghiem nay cd the triing nhau.
Cdch gidi 2 Do hai va' ciia phuang trinh \ax + b\ = \cx + d\ ludn khdng am nan khi binh phuang hai va' cua nd, ta dugc phuang trinh tuong duong. Nhu vay cd the giai phuang trinh nau d vf du 1 nhu sau : ( l ) c ^ (mx-2)^.= (x + m)^ <=> (m^ - l)x^ - 6mx + 4 - m^ = 0. GV: Hudng ddn thuc hien\\{2\vd thuc Men hoqt ddng ndy trong 4 phut.
293
Hoat ddng cua GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Khi m = 1, phuang trinh trd -6x + 3 = 0 hay x = — thanh : 2 ^ c a u hdi 2 Ggi y tra Idi cau hoi 2 Khi m = - 1 phuang trinh trd 6x = -3 c^x=—. thanh : 2 c a u hdi 3 Ggi y tra Idi cau hoi 3 Khi m ^ ±1, phuang trinh cd -m + 2 , m + 2 nghiem la : va m+1 m-1 Hai nghiem nay cd the triing nhau. HOAT DONG 2 2.
Phuang trinh chiia an d miu thurc
Khi giai phuang trinh chiia in d miu thiic, ta phai chii y den diiu kien xac dinh ciia phuang trinh. GV: Hudng ddn HS gidi vi du 2 bang cdch neu vdn de: HI. Tim diiu kien xac dinh cua phuong trinh. H2. Hay dua phuong trinh vl dang phuong trinh bae nhit. H3. Giai va bien luan phuang trinh, chii y den diiu kien. GV: Hudng ddn HS gidi vi du 3 bdng cdch neu van de sau (tuang tu nhu phdn neu vdn de khi gidi vi du 2) : GV: Hudng ddn thuc hien\H3\vd thuc Men hoqt ddng ndy trong 4 phut. Hoat ddng cua GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Diiu kian ciia phuang trinh Diiu kien cua phuang trinh la x > a la gi? 294
c a u hdi 2 Hay giai phuang trinh tren.
Ggi y tra Idi cau hdi 2 (x^ + 4x + 3)Vx-a =0 'x-a = 0 <=>
x^+4x + 3 = 0 X= a
<»
x = -3 x= -l
cau hdi 3 Ka't luan.
Ggi y tra Idi cau hoi 3 Chgn (B).
295
TOM TAT BAI HOC 1.
Phuong trinh dang | ax + b | = | ex + d | Cdch gidi 1 Chiing ta da biet: | X | = | Y | <=>X = ±Y(vdiXvaYlasdtuyy). Tuong tu, ta cd : I ax + b I = I CX + d I <=> ax + b = ± (ex + d). Nhu vay, mudn giai phuong trinh | ax + b | = | ex + d |, ta chi viec giai hai phuang trinh ax + b = ex + d va ax + b = -(ex + d) roi la'y tat ca cac nghiem thu dugc. Cdch gidi 2 Do hai ve ciia phuang trinh | ax + b | = | ex + d | ludn khdng am ndn khi binh phuang hai va' cua nd, ta dugc phuang trinh tuong duong.
2.
Khi giai phuang trinh chiia in d miu thiic, ta phai chii y da'n diiu kien xac dinh cua phuang trinh, gom nhiing x lam cho miu thiic khac 0.
HOAT DONG 3
HUONG DAN BAI TAP SACH GIAO KHOA Bai 22. Hudng ddn cdu a) Hoat ddng cua GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Dieu kien cua phuang trinh Diiu kien ciia phuang trinh la la gi?
-4
296
c a u hdi 2 Hay giai phuang trinh tran.
Ggi y tra Idi cau hdi 2 2 ( x ^ - l ) = 2 ( 2 x + l ) - ( x + 2) o
2x^ - 3x - 2 = 0, X = 4 (loai gia tri — ) .
Trd Idi b) h) Diiu kian : x ^ \ va x ^ — Khi dd ta cd : 3 2x-5 ^ 5x-^ ^ x-1 3x + 5
^2x _ 5)(3x + 5) = (5x - 3)(x - 1)
o x^ + 3x - 28 = 0. Phuang trinh nay cd hai nghiem x = 4 va x = - 7 ; ca hai diu thoa man diiu kien tran nan diu la nghiem cua phuang trinh da cho.
Bai 23. Hudng ddn cdu b) Hoat ddng cua GV
Hoat ddng ciia HS
cau hdi 1 Ggi y tra Idi cau hdi 1 Diiu kian cua phuang trinh Diiu kien ciia phuang trinh la la gi? X7^4. c a u hdi 2 Ggi y tra Idi cau hdi 2 Hay giai phuang trinh tren Khi m =i^ 3, vdi diiu kien tran, phuong khi m 7^ 3. trinh da cho tuong duong vdi: — ^ = m + 2 (*). x-4 Tir dd, neu m = - 2 thi phuong trinh (*) vd nghiem, keo theo phuang trinh da cho vd nghiem; neu m ^ -2 thi phuong trinh (*) cd
297
1
4m+ 9 +4= m+2 m+2 , 4m + 9 ^3 Xet diiu kian x ^ 3, ta cd m+2 <^m ^ - 3 . Ka't luan (vdi gia thie't vn^3): m = - 2 hoac m = - 3 : phuong trinh da cho vd nghiem. m ^ - 2 va m 9^ - 3 : phuang trinh da 4m + 9 chocd nghiem x = m+2
nghiem la x =
Trd Idi a) a) Khi m = 3, da thay phuang trinh nghiem diing vdi mgi x ^ 4.
Bai 24. Hitdng ddn cdu a) Hoat ddng cua GV
Hoat ddng ciia HS
cau hdi 1
Ggi y tra Idi cau hdi 1 Dieu kian cua phuang trinh Dieu kien cua phuang trinh la la gi? V x e R.
c a u hdi 2 Hay giai phuang trinh tren.
Ggi y tra Idi cau hdi 2 1 2ax + 3 1 = 5 c:> 2ax + 3 = ± 5 <:^ 2ax = 2 hoac 2ax = - 8 . - Khi a = 0, phuang trinh vd nghiem. - Khi a 9!: 0, phuang trinh cd hai ... 1 . 4 nghiem x = —va x = — a a
Trd Idi b) b) Diiu kien : x T^^ ± 1. Khi dd phuong trinh da eho tuong duong vdi 298
phuang trinh f(x) = 0, trong dd f(x) = x^ - 2mx + m^ - m + 1 = 0
(*)
Da tha'y (*) la phuang trinh bae hai vdi biet thiic A = m - 1. Do dd vdi m > 1, nd cd nghiem la Xl = m - V m - 1 va X2 = m + Vm-1 2 Phuang trinh (*) nhan x = 1 lam nghiem neu f(l) = m - 3m + 2 = 0, tiic la m e {1; 2}. Cu the hon, neu m = 1 thi Xi = X2 = 1; na'u m = 2 thi
Xl = 1 nhung X2 = 3 ^ 1. 2
Tuy nhien, f(-.l) = m + m + 2 ?=: 0 vdi mgi khdng bao gid nhan - 1 lam nghiem.
m nan phuong trinh (*)
Ta cd ka't luan sau : Vdi m < 1, phuong trinh da cho vd nghiem; Vdi m = 2, phuang trinh cd nghiem duy nha't x = 3. Vdi 1 < m 5^ 2, phuang trinh cd hai nghiem Xl = m - V m - 1 va X2 = m + Vm-1 ;
299
Luyen tap (tiet 8) I.
MUC TIEU
1.
Kien thirc On tap lai toan bd nhirng dang phuong trinh vita hgc.
2.
Kl nang On tap each giai phuong trinh quy vl bae nhat bae hai.
3.
Thai do Tu tin, ti mi trong thuc hanh giai phuong trinh. Phat trien tu duy trong qua trinh giai va bien luan phuong trinh.
II.
CHUAN BI CUA GV VA HS
1.
Chuan hi ciia GV : Chuin bi bai kl cac bai tap. Chuan bi mot so hinh ve trong SGK, phan mau,...
2.
Chuan bi cua HS : Cin dn lai mdt so kid'n thiic trong 3, xem lai cac \H\ va cac vf du. Dgc va lam bai trudc d nha.
IIL P H A N P H O I THCII LUONG Bai nay 1 tiet : IV.
TIEN TRINH DAY HOC
A. Bai cu Cau hoi 1 Nau each giai va bien luan phuong trinh cd chiia in d miu. cau hoi 2 Neu cac budc bien luan phuong trinh cd chiia can thiic.
300
B. Bai mdi HOATDONGI Bai 25. GV: Chita cdu a) Hoat ddng ciia HS
Hoat ddng cua GV
cau hdi 1
Ggi y tra Idi cau hdi 1 Phuang trinh tran tuong duong Binh phuang hai ve ta dugc vdi hai phuang trinh nao? (mx + 3)[9m - 2)x - 1] = 0.
Cau hdi 2 Hay giai phuang trinh tren.
Ggi y tra Idi cau hdi 2 - Khi m = 0, phuang trinh cd mdt nghiem x =
= — m-2 2 - Khi m = 2, phuong trinh cd mdt 3 3 nghiam x = = m 2 - Khi m ^ 0 va m ^ 2, phuang va trinh cd hai nghiem x = m-2 3 X =
m Trd Idi cdc cdu con Iqi: b) - Vdi a = 0, phuong trinh cd nghiem x = a + 1 = 1; - Vdi a = 1, phuong trinh cd nghiem x = 2(a + 1) = 4; - Vdi a ^ 0 va a Tt 1, phuong trinh cd hai nghiem la Xl =2(a+ 1) vax2 = a+ 1. Ggi y : Vdi diiu kien x 9^ 2 va x ;t 2a ta cd
301
^ + ^ =1 x-2 x-2a o
a(x - 2a) + X - 2 = (x - 2)(x - 2a)
< : : > x ^ - 3 ( a + l ) x + 2 ( a + 1 ) = 0 (*) Phuang trinh (*) ludn cd 2 nghiem Xi = 2(a + 1) va X2 =: a + 1. Xet cac diiu kien : Xi^2<=>2a + 2 ? t 2 < » a 9 t 0 ;
X2 7!:2 a + 1 *2
<»
a^\
Xl ^ 2a <» 2a + 2 9^ 2a (vdi mgi a); X2 ^ 2a<:e> a + 1 i^2a < » a^tl 3 m +4 c ) - Vdi m ^ l v a m ^ — . phuang trinh cd n ghiam X = m- 1 -
3 Vdi m = 1 hoac m = — , phuang trinh vd nghiem.
Ggi y: DKXD:x;>^-l.Khidd: mx-m-3
, . ,x . = l o ( m - l)x = m + 4
(1)
x+1 Vdi m = 1, da tha'y (1) vd nghiem. Vdi
1
m^\,
/IN
"^ + 4
(l)<::i>X=
m-1 Xet diiu kien x ;^ - 1 , ta cd : "1 + 4 m-1
, ^ , . 3 = - 1 <» m + 4 = -m-+ 1 <^ m = - 2
Do dd na'u m = — thi gia tri x = bi loai va phuong trinh vd nghiem. 2 ^ • m-1 • • d ) - Vdi k = - 3 hoac k = - 9 , phuong trinh cd nghiem x = 0; Vdi k ;^ - 3 va k 7!: - 9 , phuang trinh cd hai nghiem la X = 0 va X = - ( k + 6).
302
Ggi y : Vdi diiu kien x ;^ ±3, ta cd phuong trinh tuong duong x^ + (k + 6)x = 0. Phuong trinh nay co hai nghiem la x = 0 va x = -(k + 6). Tuy nhian. diiu kien tran se loai bd nghiem thir hai khi k e {-3; - 9 } . HOATDONG 2 Bai 26. GV: Chaa cdu c) Hoat ddng ciia GV cau hdi 1
Hoat dong ciia HS Ggi y tra Idi cau hoi 1
Tim diiu kien xac dinh cua DKXD ciia phuang trinh la : x > 1. phuang trinh. cau hdi 2 Hay giai phuang trinh trdn.
Ggi y tra Idi cau hdi 2 Phuong trinh tran ludn cd nghiem x=l. Xet phuang trinh : mx + 1 = 0 +) neu m = 0, phuong trinh vd nghiem. +) neu m # 0, phuong trinh cd nghiem x = - — >l^5E±l<0o-l
Trd Idi cdc cdu con Iqi: a) (2x + m - 4)(2mx - x + m) = 0
"2x+m-4=o
a;
2 m x - x + m = 0 (2) Ta giai lin lucrt : (1)<:=> x = — ( 4 - m ) (vdi mdi m). 2 ( 2 ) « ( 2 m - l)x = - m . 1
m
1
Phuong trinh (2) vd nghiem khi m =—, cd nghiem x = khi m 5^2 l-2m 2 Tii dd ta cd ke't luan : 1 1 7 Vdi m =— phuong trinh cd nghiem x = — (4 - m) = — • Vdi m^— - phuang trinh cd hai nghiam x = — (4 - m) va x = 2 2 l-2m b) Phuang trinh cd nghiem x =
va x = m+ 1
r neu m ?!: - 1 va m ^^^ -3, m+3
cd nghiam x = — neu m = - 1 , cd nghiam x = — na'u m = - 3 . ^ • 2 2 d) Vdi diiu kien x ^^ 2, ta cd : ^ ^ = a-2c^ (a-2)(x-2) = 2a-1 x-2 c^(a-2)x = 4 a - 5
(1)
Khi a = 2, (1) vd nghiem nan phuang trinh da cho vd nghiem. Khi a ^ 2, (1) cd mdt nghiem x nay se bi loai neu :
304
4a-5 a-2
Do diiu kien x ^ 2, nghiem
4 a - 5^ = 2 < : ^ 4 a - 5 = 2 ( a - 2 ) c ^ a =1a-2 2 Ke't ludn - Khi a = 2 hoac a = - - phuong trinh vd nghiem. 1 .. 4a - 5 - Khi a ?!: 2 va a ?t— - phuong trinh co nghiem x = 2 a—2 e) Vdi diiu kien x ^ -3, ta cd : ('m + ljx + m - 2 / ,s o / IX r, ^ = m o (m + l)x + m - 2 = m(x + 3) <=> x = 2m + 2. ' x+3 Do diiu kien x ^^^ - 3 , nghiem nay se bi loai neu 2m + 2 = -3<:^m = — Kit ludn : Phuong trinh cd nghiem x = 2m + 2 neu m^ —
vd nghiem
5 neu m = — 2 f) Hien nhian na'u a < 0 thi phuong trinh vd nghiem nen ta chi giai phuong trinh vdi gia thiet a > 0. Diiu kien ciia phuong trinh la x ;^ - 1 . Vdi diiu kien dd. "ax + l = a('x-l) ax + 1 = a «> ax + \ = -a(x-\) x-1
(\)
(2)
G i a i ( l ) . : T a c d ( l ) « 1 = - a. Phuang trinh nay vd nghiem do gia thiet a > 0. Giai (2) : Ta cd (2) <^ 2ax = a - 1. Phuong trinh nay vd nghidm khi a = 0, cd nghiam x = (tiic la a > 0). Ta edn phai xet diiu kien x^
\.
a—1 khi a ^ 0 2a
Ke't ludn : Vdi a < 0, phuong trinh da cho vd nghiem.
I)-TKBGDAIS610NC-T1
a—1 • Vdi a > 0, phuang trinh da cho cd nghiem x = 2a 305
HOATDONG 3 Bai 27. GV: Chaa cdu c) Hoat ddng cua HS
Hoat ddng cua GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 Tim diiu kien xac dinh ciia DKXD ciia phuang trinh la : x 7^ 0. phuang trinh. c a u hdi 2
Ggi y tra Idi cau hdi 2
Dat 2 X - 1 = t, dieu kien ciia
t>0
X
t la gi? c a u hdi 3 Ggi y tra Idi cau hdi 3 Hay vie't phuang trinh mdi vdi t ^ + t - 2 = 0 ^ t = l nghiem t. c a u hdi 4 Hay giai phuang trinh theo x.
Ggi y tra Idi cau hdi 4 1 x = l,x = —
Trd Idi cdc cdu con Iqi a) X =
3±4l4
Ggi y Dat y = V 4x^ - 1 2 x +11 ta cd phuong trinh y^ - 5y + 4 = 0. b ) x G {-5;-2; 1}. 2
Ggi y Dat y = | x + 2 |, ta cd phuang trinh y - 3y = 0.
306
HOATDONG 4 Bai 28. Hoat dong ciia HS
Hoat dong ciia GV
Ggi y tra Idi cau hdi 1 Cau hoi 1 Phuang trinh trdn tuang Ta cd duang vdi phuang trinh nao? |mx - 2| = |x + 4| 7m-ljx = 6 (\) <^
(m + \)x = -2
(2)
cau hdi 2
Ggi y tra Idi cau hdi 2 Phuang trinh cd nghiem duy Phuong trinh da cho cd nghiem duy nha't khi nao? nha't trong va chi trong cac trudng hgp sau day : 1) (1) cd nghiem duy nha't, (2) vd nghiem. Trudng hgp nay din den m = - 1 . 2) (1) vd nghiem, (2) cd nghiem duy nha't. Trudng hgp nay din de'n m = 1. 3) (1) va (2) diu cd nghiem (duy nha't) va hai nghiem dd triing nhau. Liic nay m 5^ ±1, nghiem cua (1) la X =
nghiem
ciia
(2)
la
m-1 2 X =
m+ 1 = <=> 6(m + 1) m-1 m+ 1 = - 2 ( m - 1) ^^ m = — 2
307
HOATDONG 5 Bai 29. Hoat ddng cua HS
Hoat ddng cua GV
Ggi y tra Idi cau hdi 1 Cau hoi 1 Phuang trinh trdn tuang Vdi diiu kien x ?^ a - 1 va x ^^^ -a - 2, duang vdi phuang trinh nao? tacd : x+1 _ X x-a+1 x+a+2 <» (x + l)(x + a + 2) = x(x - a + 1) o x^ + (a + 3)x + a + 2 = X - (a - l)x <^ 2(a+ l)x = -(a + 2) Ggi y tra Idi cau hdi 2
cau hdi 2 Phuang trinh khi nao?
(*)
vd
nghidm
Na'u a = - 1 thi (*) vd nghiem nen phuang trinh da cho vd nghiem. Ne'u a ^ -\ thi (*) cd mdt nghiem a+2 ^ ~ ~ 2fa +1 j Ta can xet xem khi nao thi gia tri nay bi loai do khdng thoa man cac diiu kien xac dinh. Ta cd (nhd ring ta dang a^-\): xet trudng hgp a+2
( D - 2fa + lj = a - l <:^-(a + 2) = 2(a + l ) ( a - l )
308
9
1
< j ^ 2 a + a = 0<=e>a = 0 hoac a = - -
(2)-^i^- = -a-2 2('a + lj
o - ( a + 2) =-2(a + 2)(a + 1) <:>(a + 2)(2a+l) = 0 <:^ a = - 2 hoac a = —1 2 Ket ludn Phuang trinh vd nghiem neu a e | - 2 ; - 1 ; — [ .
309
§4. He phu'dng^ t r i n h bae n h a t n h i e u a n (tielt 9,10,11) L
MUC TIEU
1.
Kien thufc
Nim viing khai niem phuong trinh bae nhit hai in, he hai phuong trinh bae nhit hai in va tap nghiem va y nghia hinh hgc cua chiing. Hieu ro phuong phap cdng dai so va phuong phap thd' trong viec giai he phuong trinh. Nim dugc cdng thiic giai he hai phuang trinh bae nhat hai an bing dinh thiic cap hai. 2.
KI nang
Giai thanh thao phuong trinh bae nhit hai in va cac he phuong trinh bae nha't hai in, ba in vdi he sd bing so. Lap va tfnh thanh thao cac dinh thiic ca'p hai D, D_, va D^ tii mdt he hai phuong trinh bae nhat hai in sd cho trudc. Biet each giai va bien luan he hai phuong trinh bae nha't hai in cd chiia tham sd. 3.
Thai do
Ren luyen dc tu duy Idgic va to hgp thdng qua vide giai va bien luan he phuong trinh. n.
CHUAN BI CUA GV vA HS
1.
Chuan bj cua GV : Chuin bi bai ki cac phin da hgc d ldp 9 de hudng din va dat cau hdi • Phan mau,...
310
2. Chuan bi cua HS : Cin dn lai phin phuong trinh da hgc d ldp 9 va dgc bai trudc d nha • Dgc va lam bai trudc d nha. IIL P H A N PHOI THCII LUONG Bai nay 3 tiet : Tie't thd: nhdt: tic ddu de'n hit phdn 1 Tie't thd hai: phdn 2. Tiet thd: ba : Phdn con lai vd hudng ddn gidi bdi tap IV. TIEN TRINH DAY HOC A. Bai cu cau hoi 1 Hay nau tap nghiem cua phuang trinh bae nhit 2 in. cau hoi 2 Hay tim nghiem chung ciia hai phuang trinh sau : 2x - 3y = 1 va X + 4y = 5. cau hoi 3 Nau cac phuang phap giai ha phuong trinh bae nhat hai in. B. Bai mdi HOATDONGI Nhic lai rang phuang trinh bdc nhdt hai dn (x va y) la phuong trinh dang ax + by = c (a, b va c la cac sd da cho, a^ + b^ ^^ 0)
(1)
Ta da biet phuong trinh (1) ludn cd vd sd nghiem. Trong mat phang toa do Oxy, tap nghiem ciia nd dugc bieu diin bdi mdt dudng thing, ggi la dudng thing ax + by = c. Chiing ta ciing da lam quen vdi hi phuang trinh bdc nhdt nhieu dn va each giai chiing bing phuong phap cdng dai sd hoac phudng phap the.
311
Trong bai nay chiing ta se nghian ciiu ki hon vl he phuong trinh bae nhat hai in. GV : Thuc hien thao tdc ndy trong 4 phiu. Hoat ddng cua HS
Hoat ddng cua GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 Hay lay mot vf du vl he | x - 2 y = l phuang trinh bae nhat hai in. |2x-y = l Cau hdi 2 Ggi y tra Idi cau hdi 2 Hay giai he phuang trinh tran Tir phuang trinh thii nh^t ta cd x = 1 + bang phuang phap the'. 2y, thay vao phuong trinh thii hai ta dugc : 2(l+2y)-y=l <::>3y = - l 1 <^y = — 3 1 Tir dd ta cd X = c a u hdi 3 Hay giai he tran bing phuang Ggi y tra Idi cau hoi 3 Nhan 2 ve cua phuong trinh thii nhat phap cgng dai so vdi 2 rdi trii di phuong trinh thii hai ve vdi ve ta dugc -3y=l«y=-Tii dd cd X = — 3 1.
He hai phuong trinh bae nhat hai an Cho hai phuong trinh bae nhat hai in ax + by = c va a'x + b'y = c' • Menh di chiia bien "ax + by = c va a'x + b'y = e'" dugc ggi la mdt he
hai phuang trinh bdc nhdt hai dn, kf hieu la
312
fax + by = c [a'x + b y = C Mdi cap sd (XQ; YQ) dong thdi la nghiem ciia ca hai phuong trinh trong he ggi la mdt nghiem cua he. Giai he phuong trinh la tim tap nghiem ciia nd. Cac khai niem he phuong trinh tuong duong, he phuong trinh he qua cung tuang tu nhu dd'i vdi phuong trinh. Dd'i vdi he phuong trinh, chiing ta ciing cd nhirng phep bien doi tuong duong, tiic la bien doi mdt he phuong trinh thanh mdt he phuong trinh khac tuong duong vdi nd. Hai quy tic cdng dai so va quy tic thd' ma ta da hgc chfnh la nhiing phep bien doi tuong duang cac he phuang trinh. GV : Hudng ddn thitc hien\H^\vd thitc Men hoqt dong ndy trong 4 phdt. Hoat dong cua GV Cau hoi 1
Hoat dong cua HS Ggi y tra Idi cau hdi 1
Giai he phuang trinh [2x-5y = - l [x + 3y = 5 bing phuang phap tha'
Tir phuong trinh thii hai ta cd : X = -3y + 5, tha' vao phuang trinh thii nha't ta dugc 2(-3y + 5) - 5y = - 1 <» - l i y =--11 <::>y = 1. Tir do ta cd X = 2
c a u hdi 2 Hay giai he phuang trinh [-2x + 6y = 2 lx-3y = -2
Ggi y tra Idi cau hdi 2 he vd nghiem
bing phuang phap cdng dai so c a u hdi 3 Hay giai he
Ggi y tra Idi cau hdi 3 He cd vd sd nghiem. 313
3x-y = l <
1 1 X— y= , 3 3 bing phuang phap cgng dai so
GV: Treo hinh 3.2 len bdng vd gidi thich. Gia sii (d) la dudng thing ax + by = c va (d') la dudng thing a'x + b'y = c', Khid6(h.3.1): 1) He (I) CO nghiem duy nhat
<» (d) va (d') cit nhau;
2) He (I) vd nghiem
o (d) va (d') song song vdi nhau;
3) He (I) cd vd sd nghiem
<=> (d) va (d') triing nhau.
HOATDONG 2 Giai va bien luan he hai phuang trinh bae nhat hai an a) Xdy diing cdng thicc Xet ha phuang trinh bae nhat hai in (I)
'ax + by = c a • X + b' y = c •
(\) (2)
- Nhan hai ve cua phuong trinh (1) vdi b', hai ve ciia phuong trinh (2) vdi -b roi cdng cac ve tuong iing, ta dugc (ab' - a'b)x = cb' - c'b.
(3)
- Nhan hai ve cua phuong trinh (1) vdi -a', hai ve cua phuang trinh (2) vdi a roi cdng cac ve tuong iing, ta dugc (ab' - a'b)y = ac' - a c. - Trong (3) va (4), ta dat : D = ab' - a'b, D, = cb' - c'b va D^ = ac' - ac. Khi dd, ta cd he phuong trinh he qua 314
(4)
(U)
Dx = D, Dy = Dy.
Dd'i vdi ha (II) ta xet cac trudng hgp sau : 1) D ?^ 0, liic nay he (II) cd mdt nghiem duy nha't (x;y)
D. D
Dy^
(5)
DJ
Ta tha'y day eung la nghiem cua he phuong trinh (I). GV: Hudng ddn thitc hieft\H2\ vd thuc hien thao tdc ndy trong 4 phut. Hoat ddng ciia GV
Hoat ddng cua HS
cau hdi 1 Chiing td D. Dy^ VD D thoa man phuang trinh (1) ciia he (I) (x; y) =
cau hdi 2 Chiing td D, Dy^ ^^^^^=VD D thoa man phuang trinh (2) cua ha (I) Ta chiing minh (x; y) =
Ggi y tra Idi cau hdi 1 GV cho mdt nhdm HS lam roi cu dai dien trinh bay : Ka't qua la nghiem.
Ggi y tra Idi cau hdi 2 Lam nhu tren KL : la nghiem
f D ' D
nghiem diing phuong trinh
ax + by = c. Mudn vay, ta chi cin chiing minh aD^ + bD^ = cD. That vay, thay the D, = cb' - c'b va D^ = ac' - a'c vao vd' phai ta cd : aD, + bDy = a(cb' - c'b) + b(ac' - a'c) = acb' - ba'e = e(ab' - a'b) = cD. 2) D = 0, liic nay he (II) trd thanh
315
'Ox = Dx Oy = Dy. +) Neu Dy ?t 0 hoac D, ;t 0 thi he (II) vd nghiem nen he (I) vd nghiem. +) Neu D, = Dy = 0 thi he (II) cd vd sd nghiem. Tuy nhien, mudn tim nghiem cua he (I), ta phai trd vl he (I) (do (II) ehi la he phuong trinh he qua). Theo gia thiet, hai so a va b khdng ciing bing 0 ndn ta cd the gia sit a ?^ 0 (trudng hgp b ^ 0 cung giai tuong tu). Ta cd D = ab' - a'b = 0 =^ b' = — b ; a D„ = ac' - a'c = 0 => c' =-
•c.
Bdi vay, ha (I) cd the via't thanh ax + by = c a / . , a — (ax + by) = — e. L a a Do dd, tap nghiem cua he triing vdi tap nghiem cua phuong trinh ax + by = c (ta da biet each giai phuong trinh nay). Ket qua tran cd the tdm tit nhu sau : r
<
9
ax + by = c
(a+b
a'x + b'y = c
T
^0)
fa'^ + b ' ^ ^ O j
1)D7^ 0 : He cd mdt nghiem duy nhit (x; y), trong dc Dx Dy x= -^; y = ^
D
D
2) D = 0 : +) ne'u D, ?^ 0 hoac Dy ?t 0 : He vd nghiem. +) neu D^ = Dy = 0 : He cd vd sd nghiem, tap nghiem ciia he la tap nghiem cua phuong trinh ax + by = c.
316
GV : Thuc Men thao tdc ndy trong 4 phdt. Hoat ddng cua HS Hoat ddng ciia GV c a u hdi 1 Ggi y tra Idi cau hdi 1 Hay cho mdt vi du ve he Chang ban: phuang trinh bae nha't 2 an |x-3y = 4 [2x + y = l c a u hdi 2 Hay tfnh D.
Ggi y tra Idi cau hdi 2 D=l.l-2.(-3)=7
c a u hoi 3 Hay tfnh D,.
Ggi y tra Idi cau hoi 3 D, = 4.1-l.(-3)=7
c a u hdi 4 Hay tfnh Dy.
Ggi y tra Idi cau hoi 4 D y = l . l - 2 . 4 = -7
c a u hdi 4 Hay ka't luan nghiem
Ggi y tra Idi cau hoi 4 x=l,y = -l.
b) Thicc hdnh gidi vd Men ludn Trong thuc hanh giai va bien luan he phuong trinh bae nhat hai in, dinh thiic la mdt cdng cu dem lai nhilu thuan tien. Bieu thiic pq - p'q, vdi p, q, p- q- la cac sd dugc ggi la mdt dinh thiirc cap hai va kf hieu la P P'
q (chii y each tfnh q' P ^
q'
pq' - p'q)-
Nhu vay, cac bieu thiic D, D, va D^, ma chung ta gap khi giai he (I) diu la cac dinh thiic cap hai, ching han D = ab' - a'b =
a a
b b'
317
GV: Hitdng ddn thuc Men vd thuc Men thao tdc ndy trong 4 phiu. Hoat ddng ciia GV Hoat ddng ciia HS Cau hdi 1 Hay vie't dinh thiic D^
Ggi y tra Idi cau hdi 1 e b = cb' - c'b D.= c' b'
c a u hdi 2 Hay via't dinh thiic D,
Ggi y tra Idi cau hdi 2 D.=
b
c'
= be - b'c.
GV : Hudng ddn HS ldm vi du 1. GV : Hudng ddn thuc hiin\WA\ vd thuc Men thao tdc ndy trong 4 phut. Hoat ddng ciia HS
Hoat ddng ciia GV c a u hdi 1 Hay tinh D.
Cau hdi 3 Hay tfnh D,.
Ggi y tra Idi cau hdi 1 2 -3 D= = 29; 7 4 Ggi y tra Idi cau hdi 3 D,=
1 -X
-'X
2
4
= 58
Cau hdi 4 Hay tfnh Dy.
Ggi y tra Idi cau hdi 4 2 13 = -87 Dy = 7 2
c a u hdi 4 Hay ka't luan nghiem
Goi y tra Idi cau hdi 4 _ Dx _ 58 _ ^ D 29 Dy 87 y = -^- = =-3. D 29
318
GV : Hudng ddn HS ldm vi du 2. HOAT DONG 3 3.
Vi du ve giai he phuong trinh bae nhat ba an
GV : Neu vi du 3 trong SGK vd hudng ddn cdch gidi, sau dd hudng ddn HS thuc Men | H 5 | de gidi tii'p. Cach giai vf du 3 : Tii (6) ta cd z = 2 - X - y.
(9)
Thay the z trong (7) va (8) bdi (9) : X + 2y + 3(2 - X - y) = 1 c^ 2x + y = 5; 2x + y + 3(2 - X - y) = - 1 <::> X + 2y = 7. Ta thu dugc he phuong trinh bae nha't hai in quen thudc (IV)
'2x + y = 5 x + 2y = 7-
GV : Hudng ddn thuc Men H 5 vd thuc hien thao tdc ndy trong 4 phdt. Hoat ddng ciia GV
Hoat ddng cua HS
cau hdi 1 Hay tfnh D.
Ggi y tra Idi cau hdi 1 D=3;
c a u hdi 2 Hay tfnh D^.
Ggi y tra Idi cau hdi 2 D. = 3
c a u hdi 3 Hay tfnh Dy.
Ggi y tra Idi cau hdi 3 Dy=9
c a u hdi 4 Hay ke't luan nghiem
Ggi y tra Idi cau hdi 4 X = 1; y = 3, z = -2.
319
Qua vf du tran, ta thay : Nguyen tic chung de giai cac he phuong trinh nhilu in la khic bdt an di quy vl giai eae phuong trinh hay he phuong trinh co sd an ft hon. De khu bdt in, ta ciing cd the diing cac phuong phap cdng dai sd hay phuong phap the gidng nhu dd'i vdi he phuong trinh hai in. GV : Hudng ddn thitc hiin\H6\ vd thuc Men thao tdc ndy trong 4 phiu. Hoat ddng ciia GV
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 c a u hdi 1 Riit X tii phuang trinh thii 3 X = 2y +4z + 1 tha' vao hai phuang trinh dau. |7y + 3 z = l l | 6 y + 13z = - l c a u hdi 2 Hay ka't luan nghiem
320
Ggi y tra Idi cau hdi 2 (x;y;z) = ( l ; 2 ; - l ) .
TOM T A T
BAI H O C
'
ax + by = c a'x + b'y = c
l)D9i
fa^+b^^Oj (a'^+b'^^Oj
0 : He cd mdt nghiem duy nhit (x; y), trong dd
Dx Dy x= - ^ ; y = - ^ D D
2) D = 0 : +) neu Dx ^ 0 hoac Dy T^ 0 : He vd nghiem. +) na'u D^ = Dy = 0 : He cd vd sd nghiem, tap nghiem ciia he la tap nghiem cua phuang trinh ax + by = c.
HOATDONG 4 HU6NG DAN BAI T A P SGK Bai 30. Hoat dong ciia GV cau hdi 1
Hoat ddng cua HS Ggi y tra Idi cau hdi 1
Ggi tap nghiem ciia phuang trinh S ' = { ( x ; y ) l x € R , y G R } . (1) la S, hay xac dinh tap nghiem ciia phuang trinh (2)? c a u hdi 2 Nghiem ciia he la gi? c a u hdi 3 Ke't luan su lua chgn.
21-TKBGOAIS610NC-T1
Ggi y tra Idi cau hdi 2 S n S' = S
Ggi y tra Idi cau hdi 3 Chgn (C).
321
Bai 31. Hoat dong cua GV
Hoat ddng ciia HS
Cau hoi 1 Tfnh D, D, va Dy.
Ggi y tra Idi cau hdi 1
Cau hdi 2
Ggi y tra Idi cau hoi 2
D = - 1 7 ; D , = 5,Dy= 19.
Nghiem ciia he la gi?
_
5 _ 17'^~
19 17'
Tra Idi b)
b)(V3,-2V2). Ggi y : Ta cd : D = - 1 ; D^ = -V3 ;Dy = 2V2
Bai 32. GV: Hudng ddn cdu a) Hoat ddng ciia GV Cau hdi 1
Ggi y tra Idi cau hdi 1
1 1 Dat X = - . Y = - ta duac ha nao? X
y
Cau hdi 2
|3X-4Y = -1
D = - 3 0 ; D , = - 1 0 , Dy = - 1 5 .
c a u hdi 3
Ggi y tra Idi cau hdi 3
Nghiem ciia he la gi?
Trd Idi b): b) (x; y) =
J6X + 2Y = 3
Ggi y tra Idi cau hdi 2
Tfnh D, D, va Dy.
f
I 322
Hoat ddng cua HS
5 )
X ; — X
2 )
vdi X e R.
Giai he tran, dugc (X, Y) = (I; 1). Tir dd suy ra nghiem cua he ban diu.
Ggi y : Vdi diiu kien
[3(x + y) •
^
x-y 5x-y_5 y-x 3
x^y,tac6:
^ ^
r3(x + yj = - 7 f x - y j | 3 r 5 x - y j = 5fy-x)
fl0x-4y = 0 [20x-8y = 0
Tap nghiem ciia he phuang trinh nay triing vdi tap nghiem cua phuang trinh : xeR 10x-4y = 0 o 5 x - 2 y = 0 o 5
[y-x Diiu kien x ;^ y thoa man khi va chi khi x^O.
Bai 33. GV: Hudng ddn cdu a) Hoat ddng cua HS
Hoat ddng cua GV c a u hdi 1 Tfnh D, D, va Dy.
Ggi y tra Idi cau hdi 1 D = m ^ - 1 ; D, = m ( m + l ) ; Dy = m + 1
Cau hoi 2 Hay bien luan nghiem cua he.
Ggi y tra Idi cau hoi 2 - Na'u m 1^ ±1 thi ha cd mdt nghiem
r m . 1 ^ 9
^m-1
m-ly
- Na'u m = 1 thi ha vd nghiem - Ne'u m = - 1 thi ha cd vd so nghiem tfnh theo cdng thirc
JxeR [y = - x
323
Trd Idi b): 5(a + l)^
b) - Na'u a 9^ - 3 thi ha cd mdt nghiam V a+3
a+3
}
- Na'u a = -3 thi he vd nghiem. Ggi y
D = -(a + 3); D, = 5; Dy = -5(a + 1)
Hudng ddn hay ddp sd cdc bdi tap con Iqi: Bai 34. (x; y; z) = (4; 5; 2). Ggi y : Trir tiing ve phuang trinh thii nhat vao hai phuang trinh sau thi dugc mot he hai phuang trinh bae nhit hai an. Bai 35. Ij, I2 va I3 la nghiem cua he phuong trinh 'Il-l2-l3=0 RlIl+R2l2=U .R2I2 "^313 =0Il « 1,33A; I2 « 0,74A; I3 « 0,59A.
324
MOT S6 B A I T A P T R A C NGHlfiM 1.
Cho phuang trinh
x + 3y = 7
(1)
Cap nao sau day la nghiem ciia (1)?
2.
(a)(l;l);
(b)(l;-2);
(c)(l;3);
(d) (1; 2)
Phuang trinh x + 2y = 1 (a) cd mdt nghiem 0; (b) cd hai nghiem
2j va(l;0);
0;
(c) cd vd so nghidm
1-x.
(d) ca ba ka't luan diu sai. Cho phuang trinh 3x - 4y = 5 cd nghiem (x^,; y^). Ggi d la dudng thing cd phuang trinh : 3x - 4y = 5. Khi dd (a) M(x^; y^) e
d;
(b) M(-x^; yo) e d;
(c) M(Xo; -y^) e d;
(d) M(-x^; -y^) G d;
Hay chgn kit qua diing. 4.
Ha phuong trinh f2x + 3y = 4
l-x +y =2 cd nghiam la : (a)
i
2 _S_]
5'
5 J'
(b)
2 ^ 5'5j
325
(c)
5.
8\ 5>
(d)
2 8^ 5'5
He phuang trinh |aix + 3bi=cia) [a2X + b2y = C2(2) Dat
D = (aib2 - a2bi); D^ = Cib2 - C2bj; Dy = C1C2 + Cia2
Khi dd : Neu D 9^ 0 thi (a) He CO nghiem duy nhat (b) He cd vd sd nghiem (c) He vd nghiem (d) ca ba ka't luan diu sai. 6.
He sau day cd nghiem duy nhat [mx + y = m [x + my = m khi
7.
(a) m ?!: 1;
(b) m ^-I;
(c)m?i±l;
(d)m = ±l.
Cho ha phuang trinh fmx + y = m [x + my = m He cd nghiem khi
8.
(a)m?i 1;
(b)m9i-l;
(c)m7i±l;
(d)m?iO.
Cho ba dudng thing : dj : 2x + 3y = 1; d2 : x - y = 2; d3 : mx + (2m + l)y = 2. Ba dudng thing tran dong quy khi
326
(a)m = 13;
(b) m = 12;
(c)m = 14;
(d)m=15.
9.
He phuong trinh x+2y+z=l 2x-y-z =— 2 -3x + y - z = - l cd nghiem la (a) (c)
22'
ir22y
'__9_ __3_. , 22' l l '
1_^ 22
(b)
_9_ J_ _ ] _ i, 2 2 ' i r 2 2
(d)
1 f 9 22 11 22
Ddp dn: 1. (d).
2.(c)
3. (a).
4. (b).
5. (a)
6. (c).
7. (b).
8. (b).
9. (d).
BAITAPTUGIAI 10.
Cho phuong trinh 2x + y = 3. Cap sd nao sau day la nghiem cua phuang trinh?
11.
(a)(0;-l);
(b) (1; 1);
(c)(l;0);
( d ) ( - l ; 1)
Cho phuong trinh x + (m - l)y = 3 (1) (a) Vdi m = 1 thi dudng thing x = 3 la bieu diin nghiem ciia (1) DDiing
DSai
(b) Vdi m = 0 thi dudng thing y = x - 3 la bieu diin nghiem cua (1) DDiing
DSai 327
(c) Vdi m = - 1 thi dudng thing y = - - + 3 la bieu diin nghiam ciia (1) 2 DSai DDiing X 3 (d) Vdi m = 5 thi dudng thing y = -— + - la bieu diin nghiam cua (1) 4
4
DDiing Hay chgn diing - sai. 12.
fx + y = 1 Ha phuang trinh < cd nghiam la [3x + 4y = 5 (a) (1; 2);
(b) (1; 2);
(c)(-l;2);
(d)(-l;-2)
Hay chgn ket qua diing. 13.
He phuang trinh
[-x-y = l |3x + 4y = 5
,
cd nghiem la
(a) (9; -8);
(b) (-9; 8);
(c) (8; 9);
(d) (9; 8)
Hay chgn ket qua dung. 14.
f-x + y = 2 Ha phuang trinh < cd nghiam la lx + 2y = 5 (a) (c)
i1
(b)
1 3
(d)
3'3
3]
Hay chgn ket qua diing. 15.
328
['-x + 4y = 5 Ha phuong trinh < cd nghiam la -y-5x =7
Z-l 3' 3
DSai
, Jl5 18^ (a) -—; 19 U9 ( 15 18
(c)
(b) (
^^
V 19
(d)
19 19
j_8 19
' 1 5 18 V 19 19 J
Hay chgn ket qua diing. 16.
Cho ha phuang trinh fx + y = m [ x - m y = 1. He phuang trinh cd nghiem duy nhit khi
17.
(a) m = 0;
(b)m=l;
(c)m = - l ;
(d) m = 2.
He phuang trinh x+y+z=1 x-y-z=2 2x + 3 y - z = 4 cd nghiem la (a) (c)
^3 _ j _ _ 5 ^ 8' 8 V.2 ' 3_ 2
]_
8
(b) (- j. 8'" (d)
11 28
5
Luyen tap (tiet 12, 13) I.
MUC TIEU
1.
Kien thurc • Cling ed eae kien thiic da hgc vl he phuong trinh bae nhat hai in va ba in.
2.
KT nang
Giai va bien luan he hai phuong trinh bae nhat hai in cd chiia tham sd bing phuong phap tfnh dinh thiic ca'p hai; giai he ba phuong trinh bae nhit ba an (khdng chiia tham so). 3.
Thai do Cd md'i lidn he vdi hinh hgc de nhin va'n dl rdng ban. Tu tin,ti mi, cd tinh thin ddc lap cao trong hgc tap.
IL CHUAN BI CUA GV vA HS 1.
Chuan hi ciia GV : Chuan bi bai ki cac phin da hgc d Idp 9 de hudng din va dat cau hoi Pha'n mau,...
2.
Chuan bi ciia HS : On lai phin phuong trinh da hgc d ldp 9 va dgc bai trudc d nha Dgc va lam bai trudc d nha.
III. P H A N PHOI THOil LUONG Bai nay day trong 2 tiet : Tiet thd nhdt: chua cdc bdi 36 den 40 Tiet thd hai: chua cdc bdi con Iqi. IV.
TIEN TRINH DAY HOC HOATDONGI Bai 36.
330
Hoat ddng cua HS
Hoat ddng cua GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 Ggi tap nghiem ciia phuang Nghiem ciia he la S n 0 = 0 . trinh (1) la S, nghidm ciia ha -la gi? c a u hdi 2 Hay ka't luan nghiem cua he.
Ggi y tra Idi cau hdi 2 Chgn (B).
HOAT DONG 2 Bai 37. GV: Hudng ddn cdu a) Hoat ddng cua GV
Hoat ddng ciia HS
c a u hdi 1 Hay tfnh D, D„ Dy
Ggi y tra Idi cau hdi 1
D=V6+5,D,=V2-V3 Dy = - 2 .
c a u hdi 2 Hay ke't luan nghiem cua he.
Goi y tra Idi cau hdi 2
5 + 46 y= - ^
« - 0,27
5 + V6 Trd Idi b):
10
•^»l,73. 10 HOATDONG 3
Bai 38.
331
Hoat ddng cua GV
Hoat ddng ciia HS
c a u hoi 1 Ggi y tra Idi cau hoi 1 Ggi hai kich thudc (tfnh bing | x + y = p met) cua hinh chir nhat la x va [('x + 3jfy + 2j = xy + 246 y (x, y > 0). Hay lap he phuang trinh. « |-+y =p [2x + 3y = 240 c a u hdi 2 Hay ke't luan nghiem ciia he.
Ggi y tra Idi cau hdi 2 X = 240 - 2p, y =3p - 240. Diiu kien x, y > 0 trd thanh : f240-2p>0 <^ 80 < p < 120. |3p-240>0 ^
HOATDONG 4 Bai 39. GV: Chaa cdu a) Hoat ddng ciia GV
Hoat ddng cua HS
c a u hdi 1 Hay tinh D, D„ Dy
Ggi y tra Idi cau hdi 1 D = - m ( m + 3); D, = -2m(m + 3) Dy = m + 3
c a u hdi 2 Hay ke't luan nghiem ciia he.
Ggi y tra Idi cau hdi 2 Na'u m ?t 0 va m ;t - 3 thi D ;t 0 nan he cd mdt nghiam (2;
).
m Na'u m = 0 thi ha trd thanh fx + Oy = l \ , vd nghiam. [0x + 0y = 3 • Neu m = - 3 thi he trd thanh
332
|X-3y = l . . ,, ,., I co vo so nghiem [-3x + 9y = - 3 ^ • (3y + 1; y) vdiy G R . Trd Idi b): b) -Vdi m ^ t - l vam;^2 : mdt nghiem
^-m + 2 ^ m+1
m+ 4' » m+1 y
- Vdi m = - 1 : ha vd nghiem fxeR - Vdi m = 2 : ha cd vd sd nghiam tfnh theo cdng thiic : < [y = 2 f l - x j Ggiy:D
= m^-m-2
= (m+l)(m - 2 ) ;
D, = -m^ + 4m - 4 = -(m - 2)^; Dy = m^ + 2m - 8 = (m + 4)(m - 2). HOATDONG 5 Bai 40. GV; Chaa cdu b) Hoat ddng ciia GV c a u hdi 1 Hay tfnh D, D„ Dy
c a u hdi 2 Hay ka't luan nghiem cua ha.
Hoat ddng ciia HS Ggi y tra Idi cau hdi 1 D = a^ + 6a + 5 = (a + l)(a + 5) D, = 3 a ' + 2 1 a + 3 0 . Dy= - a - 5. Ggi y tra Idi cau hdi 2 1) He cd nghiem duy nhit, tiic la D ?i: 0 (xay ra khi va chi khi a T^ - 1 va a ^ -5). 2) He cd vd sd nghiem, tiic la D = D, = D^ = 0 (xet cu the vdi a = - 1 va 333
a = -5. Ket qua chi cd a = - 5 la thoa man. Na'u m = - 3 thi he trd thanh fx-3y-l . , ,. i CO vo so [-3x + 9y = - 3 ' nghiem (3y + 1; y) vdi y G M. Trd Idi y a): a)a^O. 2
Ggi y : Tfnh dugc D = a He cd nghiem trong hai trudng hgp sau : 1) He cd nghiem duy nha't, tiic la D T^^ 0 (xay ra khi va chi khi a ^ 0). 2) He cd vd sd nghiem, tiic la D = D^ = Dy = 0 (khdng xay ra). HOATDONG 6 Bai 41. GV: Chaa cdu b) Hoat ddng ciia GV
Hoat ddng ciia HS
Cau hdi 1 Hay tinh D, D„ Dy
Ggi y tra Idi cau hdi 1 D = ab - 6; D, = 2b - 4; Dy= 4a - 12.
c a u hdi 2 Hay ka't luan nghiem cua he.
Ggi y tra Idi cau hoi 2 Na'u ha phuong trinh da cho vd nghiem thi D = ab - 6 = 0. Cd 8 cap sd nguydn (a; b) thoa man diiu kian nay la (l;6),(-l;-6),(6;l),(-6;-l),(2;3). (-2; -3), (3; 2) va (-3; -2). Trong so dd, chi cd cap (a; b) = (3; 2) la khdng thoa man diiu kien cua bai toan (liie nay he phuong trinh ed vd sd nghiem). Vay cd 7 cap thoa
334
man yau cau ciia de bai. HOATDONG 7 Hudng ddn hoac ddp sd cdc bdi tap con Iqi: Bai 42. Xet ha phuong trinh fx + my = 3 [mx + 4y = 6 ta CO :
D = 4 - m^; D, = 12 - 6m = 6(2 - m); Dy = 6 - 3m = 3(2 - m). a) (dl) va (dj) cit nhau
o
D 7 ^ 0 < : ^ 4 - m ^ ^ 0 <^ m ^ ± 2 .
b) (dj) // (d2)
o D = 0 va D, 9^ 0 (hoac Dy ^ 0) <:^ m = - 2.
c) (dj) va (d2) triing nhau
c^ D = D,, = D, = 0 o
m = 2.
Bai 43. Ddp sd: (x; y; z) = (4; 2; 5).
Bai 44, Hudng ddn: Doi : 1,5 trieu dong = 1500 nghin dong; 2 trieu ddng = 2 000 nghin dong; 1200 dong = 1,2 nghin dong; 1000 dong = 1 nghin dong. a) Dl thiy f(x) = 1 500 + 1,2x5 g(x) = 2000 + x. b) GV tu ve.
335
c) Hoanh do giao diem M ciia hai do thi la nghiem ciia phuang trinh f(x) = g(x), tiic la 1 500+ l,2x = 2000 + x. Phuang trinh nay cd nghiem duy nha't la x = 2500. Tung do ciia M la g(2 500) = 4 500. vay toa do ciia M la (4 500; 2 500). Ta tha'y khi x > 2500 thi dudng thing y = f(x) d phfa tran dudng thang y = g(x). Tir dd suy ra y nghia kinh ta' cua dilm M nhu sau : Na'u dung diing 2 500 gid bom thi sd tiln phai tra (tiln dien va tien may) cho hai may bam la nhu nhau (va bing 4 500 nghin ddng); Na'u diing ft ban 2 500 gid bom thi mua may thii nha't se tiat kiem han. Na'u diing nhilu ban 2 500 gid bom thi mua may thii hai tilt kiem hon.
336
§5. Mot so' vi du ve he phtfofng^ trinh bae hai hai an (tiet 14) L
MUC TIEU
1.
Kien thurc
Nim dugc cac phuang phap chu yeu giai he phuang trinh bae hai hai in, nhit la ha phuang trinh ddi xiing. 2.
KT nang
Bia't each giai mdt sd dang he phuong trinh bae hai hai an, dac biet la cac he gdm mdt phuang trinh bae nhit va mdt phuong trinh bae hai, he phuong trinh dd'i xiing. Bia't each giai va bien luan ha hai phuang trinh bae nhit hai in cd chira tham so. 3.
Thai do Cd each nhin rdng ban vl he phuong trinh. • Tu tin trong thuc hanh. Lian- ha nhilu vdi thuc tiln.
U.
CHUAN BI CUA GV VA HS
1.
Chuan bi ciia GV : Chuin bi bai ki cac phin da hgc de hudng din va dat cau hdi Pha'n mau,...
2.
Chuan bi cua HS : Cin dn lai phin dinh If Vi-et. Dgc va lam bai trudc d nha.
IIL P H A N PHOI T H d l LUONG Bai nay day trong 1 tiet:
22-TKBGOAIS610NC-T1
337
IV.
TIEN TRINH DAY HOC
A. Bai cu Cau hoi 1 Hay neu dinh If Vi-et va cac iing dung da hoc. Cau hdi 2 Tim hai so x va y bie't x + y = 3, x.y = 2. B. Bai mdi HOATDONGI De giai mdt he phuang trinh bae hai vdi hai in, ta ciing thudng diing cac phuong phap quen thudc nhu phuong phap thd', phuang phap cdng dai sd va phuang phap dat in phu. Tat nhian, viec chgn phuang phap nao con phu thuoc vao cac phuong trinh cu the. Sau day la mdt sd vf du don gian.
Vidul. Giai he phuong trinh : 'x + 2y = 5 (I)
x ^ + 2 y ^ - 2 x y = 5,
Cdch gidi Diing phuang phap the', tfnh x theo y tir phuang trinh thii nhit roi the vao phuong trinh thu hai, se dugc : 'x = 5 - 2 y (la)
10y^-30y + 20 = 0.
GV: Hitdng ddn HS thuc Men H I vd thuc Men hoqt ddng ndy trong 4 phiit. Hoat ddng cua GV
Hoat ddng ciia HS
Cau hdi 1 Ggi y tra Idi cau hdi 1 Tii phuang trinh thii hai hay Ta cd y = 1 hoac y = 2 tim y.
338
Cau hdi 2 Hay ka't luan nghiem cua ha.
Ggi y tra Idi cau hdi 2 |'x = 5 - 2 y fx = 5 - 2 y (la) <» hoac \ ^
ly = l
• ly= 2
rx = 3 _ rx = l <^ { hoac
He (I) cd hai nghiem (3; 1) va (1; 2). Vi du 2. Giai he phuang trinh : (II)
X +xy + y^ = 4 xy + x + y = 2.
Cdch gidi Ta cd nhan xet ring mdi phuang trinh trong he da cho la mdt bieu thiic dd'i xiing dd'i vdi X va y (nghia la : khi thay thi x bdi y va y bdi x thi bilu thirc khdng thay doi). Trong trudng hgp nay, ta ddng each dat in phu : S = X + y va P = xy. Khi do, x^ + xy + y^ = (x+y)^ - xy = S^ - P. Do do, tir he (II), ta cd he phuang trinh (in la S va P) S2-P = 4 S + P = 2. Dl tha'y he nay cd hai nghiem la S = -3 fS = 2 va P =5 P=0 Dodd (II) ^(Ua)r^y
= -' lxy = 5
hoac(IIb) l ' - ' ^ ^ ' • lxy = 0
GV: Hudng ddn HS thuc Men[H2| vd thuc Men hoqt ddng ndy trong 4phiu. 339
Hoat ddng ciia GV Cau hdi 1
Hoat ddng ciia HS Ggi y tra Idi cau hdi 1 He cho tha'y x va y la hai nghiem cua phuang trinh bae hai
Giai ha < • lxy = 5
f^ + 3t + 5 = 0. Phuang trinh nay vd nghiem ndn ha (Ila) vd nghiem.
c a u hdi 2 fx + y = 2 Giai he < • [xy = 0
Ggi y tra Idi cau hdi 2 Dl tha'y ha (lib) cd hai nghiam la (0; 2) va (2; 0).
c a u hdi 3 Ka't luan nghiem ciia phuang Ggi y tra Idi cau hoi 3 trinh da cho. (0; 2) va (2; 0). Vi du 3. Giai ha phuang trinh : (HI)
X -2x = y y 2 - 2 y = x.
Cach giai Ta cd nhan xet : Trong he (III), neu thay the dong thdi x bdi y va y bdi x thi phuang trinh thii nha't bien thanh phuang trinh thii hai va ngugc lai, phuong trinh thii hai bien thanh phuang trinh thii nha't. Dd'i vdi he phuang trinh cd tfnh chit dd, ta thudng giai bing each trir tiing ve hai phuong trinh trong he. Cu the, dd'i vdi he (III) ta cd : Trur titng ve hai phuang trinh trong he, ta dugc : (x^ - y^) - 2(x - y) = -(x - y) <=> ( x - y ) ( x + y - 1) = 0 <» X - y = 0 hoac x + y - 1 = 0. Dodd: 340
'x-y = 0 (Ill)
«(IIIa)
.^-2x = y
hoac (Illb)
x+y-l=0 x ^ - 2 x = y.
Ta chi edn phai giai hai he (Illa) va (Illb) ma ta da biet each giai. GV : Hudng ddn HS thuc hien\HZ\ vd thuc hien thao tdc ndy trong 4 phiit. Hoat ddng cua GV c a u hdi 1 rx-y = 0 Giai he
X -2x = y
c a u hdi 2 fx + y - l ^ O Giai ha < . [ x ^ - 2 x = y.
Hoat ddng cua HS Ggi y tra Idi cau hdi 1 He nay cd hai nghiem (0; 0) va (3; 3)
Ggi y tra Idi cau hdi 2 ha nay vd nghiem.
c a u hdi 3 Ggi y tra Idi cau hdi 3 Ka't luan nghiam ciia phuang (0;0)va(3;3) trinh da cho. Chu y: 1) Cac he phuong trinh cd tfnh chit nhu trong hai vf du 2 va 3 dugc ggi chung la hi phuang trinh ddi xdng (dd'i vdi hai in). 2) Ta cd nhan xet sau day : Ni'u mot hi phuang trinh ddi xung cd nghiem Id (a; b) thi nd cung cd nghiem Id (b; a). Nhan xet dd rat hiiu fch khi gap cac bai toan vl he phuong trinh ddi xiing. GV: Hudng ddn HS thuc hiin\y\A\ vd thuc Men hoqt ddng ndy trong 4 phut. Hoat ddng ciia GV Cau hdi 1 Cd nhan xet gi ve he da cho.
Hoat ddng cua HS Ggi y tra Idi cau hdi 1 He nay cd tinh chat dd'i xiing nghia la ne'u (a; b) la nghiem thi (b; a) ciing la nghiem.
341
Cau hdi 2 Tim cac nghiem edn lai ciia ha
Ggi y tra Idi cau hoi 2
'3-V3 , 3 + V3^ ^
c a u hoi 3 Ke't luan nghiem ciia phuang trinh da cho.
2
'
2
Ggi y tra Idi cau hdi 3 he cd 3 nghidm
^3 + V3
va
3-V3
(0;
3-V3^ 3 + V3
TOM TAT BAI HOC Ta cd tham cac loai he phuang trinh bae 2 hai in nhu sau 1. Giai bing phuong phap the. 2. He phuang trinh dd'i xiing. HOATDONG 2
HUdNG DAN BAI TAP SGK Bai 45. GV: Hudng ddn cdu a) Hoat ddng ciia GV c a u hdi 1 Dat -y = t, ta cd he mdi nao?.
Hoat ddng cua HS Ggi y tra Idi cau hdi 1 'x + t = 2 x^+t^=164
cau hdi 2 Tim cac nghiem ciia he tren. 342
Ggi y tra Idi cau hdi 2 Dat x.t = P ta cd S = 2, P =-80. Gai ra ta dugc x= 10, t = -8
0);
c a u hdi 3 Ke't luan nghiem ciia phuang trinh da cho.
Ggi y tra Idi cau hoi 3 he (10; 8) va (-8;-10)
Trd Idi y b) : (l;-l)va
2
9
5
5 j
Bai 46. GV: Hudng ddn cdu c) Hoat ddng ciia GV
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 cau hdi 1 Trir tirng phuang trinh cua he X - y - 3x + 3y = 2y - 2x ve vdi ve, ta dugc phuang o (x-y)(x + y ) - ( x - y ) = 0 trinh nao? <^ (x-y)(x + y - 1) = 0 <» X - y = 0 hoac x + y -1 =0. Ggi y tra Idi cau hdi 2 cau hdi 2 He da cho tuang duang vdi he ( 1 ) ^ - 3 " = ^y nao? [x-y = 0 hoac (II) •
Cau hdi 3 Giai he va ket luan nghiem.
x ^ - 3 x = 2y x+y-l=0
Ggi y tra Idi cau hdi 3 Nghiem eua he la : (0;0),(5;5),(-l;2),(2;-l).
Trd Idi : a)(l;2)va(2;l). Ggi y : Dat S = x + y va P = xy.
343
b)(0;l)va(-l;0). Ggi y : Dat Xj = - x de dua vl he dd'i xiing.
Bai 47. GV: Hudng ddn cdu c) Hoat ddng ciia GV
Hoat ddng ciia HS
Ggi y tra Idi cau hdi 1 c a u hdi 1 X va y la nghiem cua phuang X ^ - S X + P = 0 trinh nao?
c a u hdi 2 Ggi y tra Idi cau hdi 2 Tim quan he giiia S va P de he A = S^-4P>0 phuang trinh sau cd nghidm. Hudng ddn cdc bdi tap edn Iqi: Bai 48 a) Ggi y : Nhan hai ve ciia phuang trinh thii hai vdi 2 rdi cdng vao phuang trinh thii nhat thi dugc (x + y)^ = 400, tiic la x + y = ± 200. Do dd he phuong trinh da cho tuong duong vdi : fx + y = 20 fx + y = - 2 0 i hoac < [xy = 96 • [xy = 96 (-8; -12), (-12; -8), (8; 12) va (12; 8). b) Binh phuong hai ve cua phuang tiinh thii hai, ta cd he phuang trinh he qua: x2-y2=55 x2y2=576 2
2
Dat u = X V = y ta cd he phuong trinh
344
u-v=55 uv = 576
Giai he phuong trinh nay bing phuong phap the vdi chii y ring u > 0 va v > 0, ta dugc u = 64; v = 9. Tii day, nghiem cua he da cho cd the liy trong bd'n cap sau : (8; 3), (8; -3), (-8; 3), (-8; -3). Thii lai, ta thiy chi cd hai cap (8; 3) va (-8; -3) la thoa man. Vay he phuang trinh da cho cd hai nghiem la (8; 3) va (-8; -3).
Bai 49. Hudng ddn: Do parabol y = f(x) cat true tung tai (0; -4) nan ham sd cd dang y = ax + bx - 4, (a ;^ 0). Ggi Xj va X2 la hai nghiem ciia phuang trinh f(x) = 0. Tii gia thiet, ta cd (Xl - X2)^ = 25 hay (X1+X2) -4xiX2 = 2 5 . tiic la
fV- -al j
- 4 - =25. a
Til dd, cung vdi diiu kien f(2) = 6, ta cd he phuang trinh vdi in la a va b sau day : '4a + 2 b - 4 = 6 b^_4c = 25 La a
hay
2a + b = 5 b 2 - 4 a c = 25a^
He phuang trinh tran cd hai nghiem (a;b) = ( l ; 3 ) v a ( a ; b ) =
-25 21
Ka't qua ta cd hai ham sd fi(x) = x X + 21
155 21; + 3x - 4
va f2(x) =
x - 4 ciing thoa man yeu cau cua bai toan. 21
345
Cau hoi va bai tap 6n t a p cYivCdng III (tiet 15, 16) L
MUC TIEU
1.
Kien thurc
On tap lai toan bd kien thiic chuong nay nhim van dung trong lam toan va la tiln dl cho cac chuong sau nay. Kiem tra chuong II va thi hgc ki 1. 2.
KT nang
Ren luyen eac ki nang giai toan da hinh thanh; bie't tong hgp cac kien thirc de giai toan va lam bai kiem tra. 3.
Thai do
Cd each nhin rdng ban, nghiem tiic trong vide hgc tap va kham pha toan hoc. IL
CHUAN BI CUA GV vA HS
1.
Chuan bi cua GV : Chuin bi bai ki cac phan dn tap chuong. Chuin bi mdt bai kiem tra 1 tiet.
2.
Chuan bi ciia HS : On tap toan bd chuong va hai chuong trudc de kem tra hgc ki.
III. P H A N PHOI THCII LUONG Bai nay day trong 2 tie't : Tiit ddu danh cho dn tap Tiet sau kiim tra 1 tiet. IV. TIEN TRINH DAY HOC B. Bai mdi 346
HOATDONGI ON TAP NHUNG KIEN THUC CAN NH6 1.
Cac phep bien doi tuang duong cac phuang trinh :
1) Thuc hien cac phep toan dai sd trong tiing ve' nhung khdng lam thay doi tap xac dinh cua phuang trinh. 2) Them vao hai ve cua phuong trinh ciing mdt bieu thiic xac dinh vdi mgi X thudc tap xac dinh cua phuong trinh (trudng hgp hay diing la quy tic chuyen ve). 3) Nhan hai ve cua phuong trinh vdi cung mdt bieu thiic xac dinh va khac 0 vdi mgi gia tri cua an thudc tap xac dinh cua phuong trinh (chii y ring chia cho mdt so tiic la nhan vdi nghich dao cua sd dd). 4) Binh phuang hai ve cua mdt phuong trinh cd hai ve ludn ciing dau vdi mgi X thudc tap xac dinh cua phuong trinh 2.
Phep bie'n doi cho phuang trinh he tjua : Binh phuong hai ve ciia mdt phuong trinh.
3.
Giai va bien luan phuang trinh dang ao: + b = 0 +) a ?^ 0 :
cd mdt nghidm duy nhat x = a
4.
+) a = 0 va b ;t 0 :
vd nghiem.
+) a = b = 0 :
nghiem diing vdi mgi x
Giai va bien luan phuang trinh bae hai mot an ajc + bjc + c = 0. Tfnh: A = b^ - 4ac, A' = b'^ - ac (vdi b = 2b') +) A < 0 (A' < 0) : vd nghiem. +) A = 0 (A' = 0) : cd mdt nghiam kep XQ =
= 2a
a
+) A > 0 (A' > 0) : cd hai nghiam phan biat Xj 2 = —
= 2a
a 347
5.
Giai va bien luan he hai phuong trinh bae nha't hai an fax + by = c
?
I a' x + b ' y = c D=
-»
= ab'-a'b; D = t
a Dy = a
i
-,
(a^ + b^ 9i 0 va a'^ + b'^ i^ 0)
b'
c c
b = cb' - c'b ; b'
c c' = ac -a c
Dy D +) D ?!: 0 : cd mdt nghiem (x; y), trong dd x =—r^ y =-r+) D = 0, D, ?^ 0 hoac Dy ^ 0 : vd nghiem. +) D = D^ = Dy = 0 : cd vd sd nghiem (x; y) tfnh theo cdng thiic -by + c X =
6.
xe (nd'u a 9^ 0) hoac < - a x + c (na'ub?tO).
Dinh Ii Vi-et (thuan va dao) : '
•
2
Hai sd Xl va X2 la hai nghidm cua phuang trinh bae hai ax + bx + c = 0 khi va chi khi chiing thoa man hai he thiic Vi-et sau : _ b _ c Xj + X2 - — ; X1X2 - — a a Dinh li Vi-et cd the dng dung de : - Nhim nghiem ciia phuong trinh bae hai. - Phan tfch mdt tam thiic bae hai thanh nhan tii : Nd'u tam thiic bae hai f(x) = ax + bx + c. cd hai nghiem Xj va X2 (cd the triing nhau) thi nd cd the phan tfch dugc thanh nhan tii nhu sau : f(x) = ax^ + bx + c = a(x - Xi)(x - X2) 348
- Tfnh gia tri cac bieu thiic dd'i xiing ciia hai nghiem ciia phuang trinh bae hai : o = xj+
b
X2 = — ;
a
_
_c
P = XjX2— —;
a
Xi^ + X2^ = S^ - 2P; xi^ + X2^ = S^ - 3PS
- Xet da'u cac nghiem ciia phuong trinh bae hai : +) Phuang trinh cd hai nghiem trai diu <^ f < 0. +) Phuang trinh cd hai nghiem duong <:^ A > 0, P > 0 va 5 > 0. +) Phuong trinh cd hai nghiem am 7.
<» A > 0, F > 0 va 5 < 0.
Giai he phuang trinh bae hai hai an
1) He phuang trinh trong dd cd mdt phuang trinh bae nha't : Dung phuang phap the. 2) He phuang trinh dd'i xiing dd'i vdi hai in x va y : Diing phuong phap dat in phu : S = x + y; P = xy. HOAT DONG 2 HUONG D A N B A I T A P SGK Bai 50. Phuang trinh dang ax + b = 0 cd the cd nghiem trong nhimg trudng hgp nao? Hudng ddn: Day la bai tap If thuyet, GV cho 1 HS tra Idi. a ^ 0 hoac a = 0, b = 0.
Bai 51. Gia sii ba phuong trinh f(x)g(x) = 0, f(x) = 0 va g(x) = 0 (vdi ciing DKXD) cd cac tap nghiem lin lugt la T, Tj va T2. Hay chgn ket luan diing trong hai kit luan sau : a) T = Tl n T2;
b) T = Ti u T2. 349
Huang ddn: Day la bai tap If thuyet, GV eho 1 HS tra Idi. Dap. Chgn b).
Bai 52. He phuang trinh dang [ax + by = c (a^ + b^ ^ 0 va a'^ + b'^ ^ 0) [a' x + b'y = c cd the cd nghiem trong nhirng trudng hgp nao? ax + y = a
Ap dung : Tim a de he phuong trinh
cd nghiem.
x + ay = 1 Huang ddn: Day la bai tap If thuyet, GV cho 1 HS tra Idi. D ^ 0, hoac D = D, = Dy = 0. '
2
3
Ap dung .• Ta cd : D = a - 1; D^ = a - I; Dy = a (1 - a). Dodd: - Na'u a ?^ ± 1 thi D ^ 0 nan he co nghilm (duy nha't). - Neu a = - 1 thi D = 0 nhung D^, ^ 0 va Dy ^^ 0 nan he vd nghiem. - Neu a = 1 thi D = D, = Dy = 0 nan he cd (vd so) nghidm. Tdm lai, he cd nghiem khi va chi khi a^
-\.
Bai 53. Biet ring phuong trinh bae hai ax + bx + c = 0 cd mdt nghiem kep XQ. Hay chgn menh di diing trong cac menh di sau : (A) Tam thiic bae hai f(x) = ax + bx + c cd the viet dudi dang binh phuong ciia mdt nhi thiic bae nha't; 2
(B) Parabol y = ax + bx + c cd dinh thudc true hoanh;
350
(C) Phuang trinh cx^ + bx + a = 0 cung cd mdt nghiem kep laXo
Hudng ddn: Day la bai tap If thuyet, GV cho 1 HS tra Idi. Chgn (B) Bai 54. Giai va bien luan phuong trinh m(mx - 1) = x + 1. Hoat ddng ciia GV
Hoat ddng cua HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Hay bie'n doi phuang trinh vl m(mx - 1) = X + 1 dang phuang trinh bae nha't. <:^ (m^- I)x = m + 1. cau hdi 2 Hay bien luan phuang trinh.
Ggi y tra Idi cau hdi 2 X=
neu m 7^: ± 1; vd nghidm m-1 neu m = 1; nghiem diing vdi mgi x neu m = -l. Bai 55. Cho phuang trinh p(x + l ) - 2 x = p + p - 4 . Tim cac gia tri cua p de a) Phuang trinh nhan 1 la nghiem; b) Phuong trinh cd nghiem; c) Phuang trinh vd nghiem. Hoat ddng ciia HS
Hoat ddng cua GV
Ggi y tra Idi cau hdi 1 Cau hdi 1 Hay bie'n doi phuang trinh ve (p-2)x = p 2 - 4 . dang phuang trinh bae nhat. c a u hdi 2 Phuang
trinh
nhan
1
Ggi y tra Idi cau hdi 2 la p = ±2. 351
nghidm khi nao? c a u hoi 3 Riuang trinh cd nghiem khi nao? c a u hdi 4 Phuang trinh vd khi nao?
Ggi y tra Idi cau hoi 3 P^2.
nghidm Ggi y tra Idi cau hoi 4 Khdng cd p.
Bai 56. Ba canh cua mdt tam giac vudng cd dd dai la ba sd tu nhidn lian tiep. Tfnh do dai cua chiing. Hudng ddn: Ggi ba canh cua tam giac l a : x - l , x v a x + l, trong dd x la so nguyan. Khi dd ta cd : (x + 1)^ = x^ + (x - 1)^ Tir dd ta cd cac canh la 3, 4, 5.
Bai 57. Cho phuong trinh (m - l)x + 2x - 1= 0. a) Giai va bien luan phuong trinh. b) Tim cac gia tri ciia m sao cho phuang trinh cd hai nghiem trai da'u. c) Tim cac gia tri ciia m sao cho tong cac binh phuang hai nghiem cua phuang trinh bang 1. Hoat ddng ciia GV c a u hdi 1 Giai va bien luan trinh.
Hoat ddng cua HS Ggi y tra Idi cau hoi 1 phuang Vd nghiem khi m < 0; x = 1 (nghiem kep) khi m = 0; x = — khi m = 1; - 1 + Vm ,, . ^ X=
khi 0 < m ^ 1.
m-1 352
c a u hdi 2 Phuang trinh cd hai nghiem Ggi y tra Idi cau hoi 2 trai da'u khi nao? K h i m - 1 >Ohaym> 1. Ggi y tra Idi cau hoi 3 c a u hdi 3 Hay tim tdng cac phuang hai nghiem.
binh
(Xi + X2)^ - 2XiX2 = I
X^+xl= i
0 ^2
-2 m-1
-2-
^
1 =1 m-1
Ggi y tra Idi cau hdi 4 c a u hdi 4 Khi nao tdng cac binh phuang hai nghiem bang 1
Ketluan :m = 2+V5
Bai 58. Vdi gia tri nao ciia a thi hai phuang trinh sau cd nghiem chung : x^ + X + a = 0 va x^ + ax + 1 = 0? Hoat ddng ciia GV
Hoat ddng ciia HS
c a u hdi 1 Ggi y tra Idi cau hdi 1 Na'u ggi nghiem chung cua Xo=l. phuang trinh la XQ, khi dd ta dugc XQ bing bao nhiau? c a u hdi 2
Ggi y tra Idi cau hdi 2 Khi XQ = 1 thi a bing bao a = -2. nhiau?
c a u hdi 3 Hay thir lai khi a = -2.
Ggi y tra Idi cau hoi 3 a = -2 thoa man.
Bai 59. Cho cac phuang trinh 23.TKBGBAISO10NC-T1
353
x^ + 3 x - m + l = 0
(1)
va
2x^ - x + 1 - 2p = 0
(2).
a) Bien luan sd nghiem ciia mdi phuong trinh bing do thi; b) Kiem tra lai ket qua tran bing phep tfnh. GV: Hudng ddn HS ldm cdc bdi tap ndy. Ve tren ciing mat phang toa do parabol y = x + 3x + 1 va dudng thang y = m •>
2
'
Ve tran cung mat phang toa do parabol y = 2x - x + 1 va dudng thang y = -2p.
Bai 60. Giai cac he phuang trinh : 2
a)
2
b)
X +y + x y = 7
2('x + y r - x y = l 2
2
X y + xy = 0 .
x^+y^-xy = 3 GV: Hudng ddn cdu a)
Hoat ddng ciia HS
Hoat ddng ciia GV
Ggi y tra Idi cau hdi 1 c a u hdi 1 He da cho tuang duang vdi he ' x 2 + y 2 = 5 nao? xy = 2 •
c a u hdi 2 Tim nghidm ciia he.
Ggi y tra Idi cau hdi 2 T = { ( l ; 2 ) , ( 2 ; l ) , ( - l ; - 2 ) , ( - 2 ; -1)}.
Trd Idi y b) : T = {(1; -1), ( - 1 ; I), ( 0 ; 4 - ) , (0; - - ^ ) , ("7-; 0), ( " T ^ i 0)} Ggi y : Dat u = X + y va V = xy, ta cd he phuong trinh in la u va v la
2 U ^ - V := 1
uv = 0
354
Bai 61. Giai va bien luan cac he phuong trinh : a)
mx + 3y = m - l 2x + fm-ljy = 3 '
b)
'5x + ('a-2jy = a ('a + 3jx + (a + 3;y = 2a.
Hudng ddn: a) - Na'u m 9i 3 va m ^ - 2 thi ha cd mdt nghiem duy nha't (x; y), trong dd: x=
m-4
1 m-3
;y
m-3
- Na'u m = 3 thi he vd nghiem. - Neu m = - 2 thi he cd vd sd nghiem (x; y) tfnh theo cdng thiic : 'xeR y = —x-1 3 b) - Nd'u a^-3vaa^l
thi he cd mdt nghiem duy nha't (x; y), trong dd a
x=y=
. a+3
- Neu a = - 3 thi he vd nghiam. - Na'u a = 7 thi he cd vd sd nghiem (x; y) tfnh theo cdng thiic : 'xeR 7 y = -x + — Bai 62. Giai va bien luan cac he phuong trinh a)
x+y = 4 xy = m
3x-2y = l b)
2
2
x^+y^ =m.
Gidi 355
a) Theo dinh If Vi-et dao, x va y la hai nghiem cua phuong trinh : z ^ - 4 z + m = 0 (1). Ta cd A = 4 - m. Do dd - Neu m > 4 thi A < 0, phuong trinh (1) vd nghiem nan he da cho vd nghiem. - Neu m = 4 thi A = 0, phuang trinh (1) cd mdt nghiem kep z = 2 nan he da cho cd mdt nghiam (x; y) = (2; 2). - Na'u m < 4 thi A > 0, phuang trinh (1) cd hai nghiem phan biet Zi 2 = 2 ± V 4 - m nan he da cho cd hai nghiem : X = 2 + V4 - m x = 2-V4^ m va < y = 2-V4-m y = 2 + 4^m b) Giai Ta cd
X2 + y 2 = m
2y = 3 x - l
2y = 3 x - l
3x-2y = l <^
<^<
4x^+4y^ =4m
4x^+f3x-l)^=4m
Xet riang phuang trinh : 4x^ + (3x -\f
= 4m <^ 13x^ - 6x - 4m + 1 = 0.
(2)
Phuang trinh (2) cd biet thiic thu ggn A' = 9 - 13(1 - 4m) = 4(13m - 1). Dodd: - Na'u m < — thi A' < 0, phuang trinh (2) vd nghiem nan ha vd nghiam. 1 3 - Neu m = — thi A' = 0, phuong trinh (2) cd mdt nghiam x = — nen ha 3 cd mdt nghiam (x; y) = ( — ; • ^ • ' "^ 13
2 ). 13
- Ne'u m < — thi A' < 0, phuang trinh (2) cd hai nghiem Xi 2 = 3±2Vl3m-l ndn ha cd hai nghiem : 13 356
(xi;yi) =
f 3-2Vl3m-l 13
va (X2; y2) =
-2-3Vl3m-l 13
f 3 + 2Vl3m-l 13
- 2 + 3Vl3m-l 13
Bai 63. Tim a, b va c de parabol y = ax + bx + c cd dinh la diem 1(1; -4) va di qua diem M(2; -3). Hay ve parabol nhan dugc. Hudng ddn: Vi diem 1(1; -4) la dinh cua parabol nan . 1=
b . . A b^-4ac . va - 4 = = ^—.hay 2a 4a 4a 2 b = - 2a va 16a = b - 4ac. Vi parabol di qua diem M(2; -3) nan - 3 = 4a + 2b + c. Ta cd ha phuang trinh (chii y rang a ^t 0) : 'b = -2a 16a = b ^ - 4 a c - 3 = 4a + 2b + c Kbit a trong he tran bing each the 2a = - b , ta dugc he phuang trinh dd'i vdi b va c : b^+8b + 2bc = 0 •8b = b^+2bc <=?• < -3 = - 2 b + 2b + c c = -3
'^b^+2b = 0
<=> <
Nd'u b = 0 thi a = 0, trai vdi diiu kian a ;t 0. Vay b = - 2 , tur dd suy ra 2
a = 1 va ham sd cin tim la y = x - 2x - 3 . Tir day, da dang ve dd thi cua nd. DS: a = l ; b = - 2 ; c = - 3 .
357
Bai 64. Cho tam giac ABC cd BC = a, CA = b va AB = c. Ta lay mdt diem M tren canh BC. Qua M, ta ke cac dudng thing ME va MF lin lugt song song vdi cac canh AC va AB (E e AB, F e AC). Hdi phai lay diem M each B mdt khoang bing bao nhiau de tong ME + MF = / (vdi / la do dai cho trudc)? Bien luan theo /, a, b va c. Hudng ddn: Dat X = MB (diiu kien : 0 < x < a). Theo dinh If Ta-let ta cd (h. 3.15):
ME X
b bx = — ^ ME = a a
MF_a-x_,,^
cfa-- x j
c a a Diiu kien ME + MF = / cho ta phuang trinh : /IN /IN /1N , bx cfa —xj /= — + — c^ (h- c)x = a(/ - c) (1) a a - Na'u b = c (tiic la giac ABC can d A) thi phuong trinh (1) vd nghiem neu / ^ c, nghiem diing vdi mgi x neu / = c. Diiu nay cd nghia la : + Khi tam giac ABC can tai A va / # AB khdng cd diem M nao trdn canh BC thoa man diiu kien ciia bai toan. + Khi tam giac ABC can tai A va / = AB thi mgi diem M tran canh BC diu thoa man diiu kien ciia bai toan. - Neu b ?!: c (tire la tam giac ABC khdng can d A) thi phuang trinh (1) co a(l — c) Xet diiu kidn 0 < x < a : mdt nghiem duy nhat x = b-c 0 < x < a c ^ O < .^iZz^ < a o b-c Vi b ?i c nan cd hai trudng hgp : + Vdi b > c, ta cd ( 2 ) < e = > 0 < / - c < b - c <:^ c < / < b . 358
0<- ^ < 1 b-c
(2)
+ Vdi b < c, ta cd : (2)c^ 0 > / - c > b - c <=> c > / > b a(l — c)
•<
Hai ka't qua tran cd nghia la gia tri x = -^ la nghiam cua bai toan (diem b-c . af/-cj M each B uigi mdt Knoang khoang cbing^^——) khi va chi khi do dai / nim giua cac do b-c dai b va c.
BAI TAP TRAC
1.
NGHIEM
Tim giao diem cua 2 dudng thing sau d i : x + 2 y = l ; d2 : 2x + 3y = - 5 (a) (3; 1);
(b) (3; -1);
(c)(-3;l);
(d)(-3;-l)
Hay chgn kit qua diing, 2.
Hoanh do giao diem ciia parabol : y = x - 2x + 5 va dudng thang d : x + y - 6 = 01a (a)
hoac 2
; 2
, -l->/5^ - I + VS (c) hoac ^ ; 3.
(b) khdng cd; , , , 1 - V 5 ^ . 1 + V5 (d) — ^ hoac — ^ .
He phuong trinh rx-2y + l = 0 [-x + 3 y - 3 = 0 cd nghiem la (a) (3; -2);
(b) (3; 2);
(c) (-3; -2);
(d) (-3; 2). 359
He phuong trinh [mx + y = m + l [ x - m y = 2006 cd nghiem duy nhit khi (a)m;tl;
(b)m^-l;
(c)m9^±l;
(d)Vm.
Phuang trinh | x - l | + 2 x - 3 = 0 cd nghiem la
-1
(a)|; 4 (c) —hoac - 2 ; 6.
hoac 2:
t
( d ) - - - hoac - 2 . 3
2 1 1 = 0 cd tap nghiem la Phuong trinh x + |x| - 2 =
(a)S={-2;l};
(b)S == { - i ; i };
(c)S={-l;2};
( d ) S := 0 .
Phuang trinh (m - 2 v 3 m - l)x + m H- V200'7m = 0
1
cd nghiem khi
8.
= V^i
(a) m ;t V3 ± 2;
(b)m
(c) m = V3 - 2;
(d)m = V3 -f -2.
Cho phuang trinh x ^ - 2 m x + m-- 2 = 0
(1)
Phuang trinh (1) cd 2 nghie m Xl, X2 thoa man 3xi 4- 2x2 = :5 khi (a)m=
-49±V4609 ^^ ;
(r\ m = - 4 9
360
(b)m Cdlm
- 4 9 H hV4609
48 - 4 9 - -^74609 48
Ddp dn: l.(b)
2.(d)
3.(d)
4. (d)
5. (a)
6.(b)
7. (b).
8. (a).
BAITAPTITGIAI 9.
Phuang trinh 2mx + x + 2 = 0 cd nghiam khi (a)m7iO;
(b)m?t--;
2 (c)m?^l;
(d)m9t2.
Hay chgn kit qua diing. 10.
Phuang trinh x - 3x - 3 = 0 cd hai nghiem |xi - X21 bing (a) 21
(b)-2I;
(c)V2T;
(d)-V2l
Hay chgn ka't qua diing. 11.
Phuang trinh X - V3x - 3 = 0 cd hai nghiem Xi va X2. 2
2 ^
Khi dd : Xl + X2 bang
(a) 6 ; (c) - 9 12.
(b) - 6 ; ;
(d) 9.
Phuang trinh x - 2x + v3 - 1 = 0 cd hai nghiem Xi va X2. Khi dd : Xl + X2 bing
(a)-14-6V3
;
(b) 14 + 6 ^ 3 ;
(c)14-6V3
;
(d)-14 + 6V3
361
Hay chgn ki't qua dung trong cdc bdi tap tic 3.37 di'n 3.40 13.
14.
15.
Ha phuang trinh (a) (6; 1);
(b) (6; -1);
(c)(-6; 1);
(d)(-6;-l).
rx-3y = 5 Ha phuong trinh { eo nghiam la [-x + 6y = 7 (a) (17; 4);
(b) (-17; 4);
(c) (-17; -4);
(d) (17; -4).
f2x-y = -12 Ha phuong trinh < cd nghiam la [-3y + 4x = 8 (a) (-22; 32); (b) (22; -32); (c) (22; 32);
362
' cd nghiam la [x-y =7
(d) (-22; -32).
GOI t DE KIEM TRA CUOI CHUONG (Thdi gian ldm bdi cho mdi de Id 40 phiit).
D e SO 1 cau 1. (5 diem) Giai cac he phuong trinh sau : 6 2 ^ - +- =3 a)
X X
xy + 4x + 4y = - 2 3 b)
y
x^+xy + y^ =19
y
cau 2. (5 diem) Cho phuong trinh ( m - l)x^ + 2 x - m + 1 = 0 a) Chiing minh ring vdi mgi m ^ \, phuong trinh ludn cd hai nghiem trai dau. b)Vdi gia tri nao ciia m thi mdt trong hai nghiem ciia phuang trinh bing -2? Khi dd hay tim nghiem kia. c) Vdi gia tri nao ciia m thi tong binh phuong hai nghiem cua phuong trinh bing 6?
Ddp dn: Cau 1. 3 2 a) Dat u = — va V = —. ha phuong trinh tro thanh X
y
f2u + v = 3 |u-2v = -l Tii phuang trinh thii hai ta cd : u = 2v - 1
(1)
363
Trong phuang trinh diu, thay the u bdi 2v - 1 ta dugc : 2 ( 2 v - l) + v = 3 <^ 5v = 5 c^ V = 1. Til do va (1) suy ra u = 1. Do dd ha da cho tuong duong vdi he phuang trinh :
x^^
fx = 3
ly vay ha phuong trinh da cho cd nghiem (x; y) = (3; 2). 2
2
2
b) Dat s = X + y va p = xy, ta cd : X + y = s - 2p. The vao he phuong trinh da cho, ta dugc he phuang trinh in s va p :
fp+ 4s = - 2 3
s 2 - 1 9 + 4s = - 2 3 ^
-7
[ s ^ - p = 19
p = s2-19
s2+4s + 4 = 0 < »
•
p = s2-19
fs = -2 <^
[p = -15 Do dd, ha phuang trinh da cho tuong duong vdi he : fx + y = - 2
{
[xy = - 1 5
rx = 3
<:> {
[y = - 5
rx = - 5
hoac <
"
ly = 3
Vay ha phuong trinh da cho cd hai nghiem (3; -5) va (-5; 3) Cau 2. a) Khi m Tt 1, ta cd a ^^ 0 nan phuong trinh da cho la mdt phuong trinh bae 2
hai. Hon niia, phuang trinh nay cd ac = -(m - 1) < 0 nan phuong trinh ludn co hai nghiem trai da'u. 364
b) Kf hieu ve trai cua phuang trinh la f(x). Phuang trinh cd nghiem x = - 2 khi va chi khi f(-2) = 0, nghia la : 7 ( m - l ) 4 - 4 - m + 1 = 0 <^ 3m - 7 = 0 <^ " ^ = 7 " Khi dd, do tfch ciia hai nghiam bing - = - 1 nan nghiem thii hai ciia a phuang trinh la —. c) Tvtn day ta di biet : vdi m T^ 1, phuang trinh ludn cd hai nghiam trai da'u. Ggi hai nghiem a'y la Xiva X2. Theo Vi-et ta cd : _
Xi+X2 -
2 m-1
_ ,
; X1X2--1.
Dodd xf + x ^ = (Xi+ X2)^ - 2 X1X2 =
•—^
+ 2;
(m -1)2 9 9
xf + X2 = 6 o
4
+2 = 6
(m-ir <=e> (m - 1)^ = 1 <::> m = 0 hoac m = 2. Tra Idi : m = 0, m = 2.
365
De so 2 cau 1. (5 diem) Giai va bien luan he phuang trinh sau (a la tham so) fax-4y = 2 [-x + ay = a - 3 cau 2. (5 diem) Cho phuong trinh x^-(k-3)x-k +6 =0
(1)
a) Khi k = - 5 , hay tim nghiem gin diing cua (1), chfnh xac den hang phin chuc. b) Tuy theo k, hay bien luan sd giao diem cua parabol y = x - (k - 3)x k + 6 vdi dudng thing y = -kx+ 4. c) Vdi gia tri nao ciia k thi phuong trinh (1) cd mdt nghiem duong?
Ddp dn: Caul. Ta cd : D=
D.
a
-4
-1
a
= a - 4 = ( a - 2 ) ( i + 2 ) ; D = 0 <::>a = ± 2 .
2 a-3
-4 = 2a + 4 ( a - 3 ) = 6 ( a - 2 ) . a
a
2 = a -3a + 2 = (a-l)(a-2) a-3
Dy = - 1
Td' dd suy ra : 1) Neu a ?!: ± 2 thi D T^ 0 va he phuong trinh cd mdt nghiem (x; y), trong dd
366
X =
D. D
6_
_ Dj^_
a + 2 ' ^~ 'D"
a-]_
a+2
2) Neu a = 2 thi D = D, = Dy = 0, he trd thanh f2x-4y = 2 | - x + 2y = - l va cd vd sd nghiem co dang tdng quat la (2y + 1; y) vdi y e M. 3) Neu a = - 2 thi D = 0, D, 5^ 0 nen he vd nghiem. Kit ludn 1) Khi a 9!: ± 2, ha cd mdt nghiem : X =
6
a+2 _ a-1 ^"a +2 2) Khi a = 2, he cd vd sd nghiem : fx = 2y + l [yeM 3) Khi a = - 2 , he vd nghiem.
cau 2. a) Vdi k = - 5 , ta cd phuong trinh bae hai x^ + 8x + 11 = 0. Phuong trinh nay cd biet thiic A' = 5 nan cd hai nghiem phan biet : Xl = - 4 - V 5 « -6,2 va X2 = - 4 + VS « - 1 , 8 . b) So giao diem ciia parabol vdi dudng thing da cho bing sd nghiem ciia ^ phuong trinh x ^ - ( k - 3 ) x - k + 6 =-kx+4
(2)
Ta cd : (2) <^ x^ + 3 x - k + 2 = 0 Phuong trinh bae hai nay cd biet thiic A = 9 - 4(-k + 2) = 4k + 1. Do dd : 367
1) Neu k < — thi D < 0, phuang trinh (2) vd nghiam nan parabol va dudng thing khdng cd dilm chung. 2) Neu k = — thi D = 0, phuang trinh (2) cd mdt nghiam nan parabol va 4 dudng thing cd mdt dilm chung. 3) Na'u k > — thi D > 0, phuang trinh (2) cd hai nghiam phan biet nan 4
parabol va dudng thing cd hai diem chung. c) Xet cac trudng hgp sau : 1) Phuang trinh <1) cd nghiam kep, nghia la A = (k -3)^ + 4(k - 6) = k^ - 2k - 15 = 0 <» k = - 3 hoac k = 5 - Neu k = - 3 thi (1) trd thanh x^ + 6x + 9 = 0, cd mdt nghiem am x = - 3 (khdng thoa man yeu ciu). - Neu k = 5 thi (1) trd thanh x^ - 2x + 1 = 0, cd mdt nghiam dudng x = 1 (thoa man yau ciu). 2) Phuang trinh (1) cd mdt nghiem x = 0, nghia la k = 6. Liic nay phuang trinh trd thanh x - 3x = 0, ngoai nghiem x = 0 edn cd mdt nghiam duong x = 3 (thoa man yau ciu). 3) Phuang trinh (1) cd mdt nghiem am va mdt nghiem duong, tiic la hai nghiem trai diu. Diiu dd xay ra khi -k + 6 < 0 hay k > 6. Tdm lai, cac gia tri cin tim cua k l a k = 5 v a k > 6 .
368
On tap hpc ki I I.
MUC TifiU
On tap va ciing cd toan bd kien thiic cua ba chuong: Menh dl tap hgp. Ham sd va Phuong trinh. Sau do lam bai kiem tra 1 tiet. n.
NOI DUNG Chuang I:
On tap vl menh dl, menh di keo theo, menh di tuong duong, menh dl phu dinh. Tap hgp, cac each cho tap hgp, cac phep toan vl tap hgp. Sai sd tuyat dd'i, chir sd dang tin. Chuang II Ham sd, tap xac dinh cua ham so, do thi cua ham sd, diem thudc do thi ham so, tfnh dong bien, nghich bien ciia ham so, ham sd chan, ham sd le. Ham sd bae nhat, do thi ciia ham sd bae nhat, he sd gdc cua dudng thing, hai dudng thing song song, ham sd y = I xl. Ham sd bae hai, tfnh dong bien nghich bien, gia tri ldn nhat, gia tri nho nha't, do thi ciia ham sd, toa do dinh, true dd'i xiing. Chuang HI Phuang trinh, diiu kien xac dinh, tap nghiem, phuang trinh tuong duong va phuong trinh he qua. Phuang trinh bae nha't, giai va bien luan phuong trinh bae nha't chiia tham so, mdt so phuang trinh dua vl bae nha't, he phuong trinh bae nha't hai in. Phuang trinh bae hai, dinh If Vi-et. Mdt sd phuong trinh dua vl bae hai. III.
Y£UCAU Hoc sinh dn tap ki ta't ca cac kiln thiic tren va van dung trong viec giai toan.
24-TKBGOAISO10NC-T1
369
MOT SO DE BAI ON TAP VA KltU
TRA HOC KI I
Gdm hai phdn: Cau hoi trie nghiem khach quan: 4 diem, lam trong 15 phiit, thu bai ngay. Cau hdi va bai tap tu luan : 6 diem
D e SO 1 I.
CAu HOI TRAC NGHlfiM K H A C H QUAN
Cau 1. Xet phuang trinh mx -2x = - x
+v7
Hay chgn ket luan sai trong cac ket luan sau: (a) Vdi m e [1, +oo) phuang trinh ludn cd nghiem. 1
/?
(b) Vdi m e [
j=—. +oo) phuang trinh ludn cd nghiem.
47
-
(c) Phuong trinh ludn cd nghidm vdi mgi m. (d) m = - 1, phuang trinh cd nghiem duy nhit. Cau 2. Ham sd y = 2x + m - 1 (a) Ludn dong bie'n trdn R. (b) Ludn nghich bien tran R. (e) Dong bia'n hoac nghich bien tan R tuy theo m. (d) Cd mdt gia tri ciia m de ham so la ham so hing. cau 3. Phuong trinh V x - 2 +— = 0 cd diiu kidn xac dinh la : X
(a) R.
(b) [2; + ^).
(c)R\{0}.
(d) R \ ( 0 ; 2 } .
Cau 4. Phuang trinh 2x + 1 = 1 - 4x tuong duong vdi phuang trinh nao sau day? ( a ) x ( x - i ) = 0; (c) X + - 7 =1- = - 71= ; Vx -Jx 370
(h)(x^+l)x
= 0;
(d) x.yJx-3
=0.
n.
CAU HOI TULUAN
cau 1. Giai va bien luan cac phuang trinh sau: a ) ( m - l)x + ( m ^ - 1) = 0; b) x^ = 2m X + m - 1. cau 2. Cho ham sd y = (m - l)x + 3x + m - 2. a) Xac dinh m de ham sd dong bien tran R. b) Vdi m 7^ 1 hay xac dinh m de ham sd cd gia tri nhd nhit. c) Vdi m tim dugc d cau b), xac dinh gia-tri nhd nhit cua ham so. Cau 3. Mdt Idp hgc cd 36 hgc sinh, trong dd cd 6 em gidi toan, 8 em gidi van, 4 ern gidi ca van ca toan. Tim sd hgc sinh khdng gidi mdn gi.
Hudng ddn vd ddp dn: I.
CAu HOI TRAC NGHIEM K H A C H QUAN Cau 1 (1 d). Chgn (c). cau 2 (1 d). Chgn (a). cau 3 (1 d). Chgn (b). cau 4 (1 d). Chgn (b).
II.
CAU HOI TULUAN cau 1. (2 d, mdi cau 1 d) a) Vdi m = 1, phuang trinh cd vd sd nghiem; Vdi m ^ 1, phuong trinh cd nghiem duy nha't x = -m - 1. b) Phuong trinh trd thanh : x^ - 2mx - m + 1 = 0. Tacd A' = m^ + m - 1. Vdi m
(-1-45
I
2
;
-1 + 45]
. _ ,^ u . u-^ , phuong trinh vo nghiem.
2 J 371
Vdi m =
hoac m =
^
phuang trinh cd nghiem kep.
z
Trudng hgp con lai phuong trinh cd hai nghiem phan biet. Cau 2. (3 d, mdi cau 1 diem) a) Ham sd ddng bien tran R khi m = 1. b) Ham sd cd gia tri nhd nha't khi m - 1 > 0 hay m > 1. N^..,.
,,
.,^, . . .
.-,. -A
-4m^+12m + l
c) Gia tri nho nhat cua ham so la — = 4a
m-1
cau 3. (1 diem). So hgc sinh gidi toan hoac gidi van la: 8 + 6 - 4 = 10 em. So hgc sinh khdng gidi mdn nao trong 2 mdn Toan hoac Van la: 36 - 16 = 20.
Deso2 CAU HOI TRAC NGHIEM KHACH QUAN cau 1. Xet phuang trinh x + 4x-3
= 4-43-x
Hay chgn kit luan diing trong cac kit luan sau: (a) Diiu kien xac dinh ciia phuong trinh la x > 3 (b) Diiu kien xac dinh cua phuang trinh la x < 3 (c) Diiu kien xac dinh cua phuang trinh la x = 3 (d) Phuang trinh cd nghiem la x = 3. cau 2. Ham sd y = x^ + 3x + 7. Hay chgn kit luan sai trong cac kit luan sau: (a) Ludn dong bien tran ( — ; + co). (b) Ludn nghich biln tran (-oo ; — ) . 372
1
3
(c) Dudng thing x = — la true dd'i xiing ciia dd thi ham so. 2 (d) Do thi ciia ham sd ludn cit true hoanh. Cau 3. Phuang trinh 4x +— = 0 cd diiu kidn xac dinh la : X
(a) x > 0 .
(b)x^O;
(c)x>0};
(d) M\{0}.
Cau 4. Phuang trinh x = 1 la phuang trinh he qua cua phuong trinh nao sau day.
IL
(a)x(x-l) = 0;
( b ) ( x ^ - l ) x = 0;
(c)lxlVx=0
(d)x(x^-l)y/x-3 =0.
CAU HOI TULUAN
Cau 1. Giai va bien luan phuang trinh sau: a) (m - l)x^ + 2x + (m + 1) = 0;
b) Ixl = 2mx + m - 1.
Cau 2. Cho ham sd y = 2x + m - 2. a) Chiing minh vdi mgi m de ham sd dong bien tran R. h) Xac dinh m de do thi ham sd cung vdi hai true tao thanh tam giac cd dien tfch bang 6. Cau 3. Giai he phuang trinh sau: rx + 2y = - l | - x + y = 2.
Hudng ddn vd ddp dn: I.
CAU HOI TRAC NGHlfiM K H A C H QUAN cau 1 (1 d). Chgn (c). 373
cau 2 (1 d). Chgn (d). cau 3(1 d). Chgn(c). cau 4 (1 d). Chgn (b). IL
CAU HOI TIT LUAN cau 1. (3 d, mdi cau 1,5 d) a) Vdi m = 1, phuang trinh da cho trd thanh 2x = 0 => x = 0; Vdi m ^ 1, phuang trinh cd A = -m < 0. Tii dd ta cd : Khi m = 0, phuang trinh cd nghiem kep x = 1, Khi m ^ 0, phuang trinh vd nghiem Kit ludn : m = 0, phuang trinh cd nghiem kep x = I; m = 1, phuang trinh cd nghiem x = 0; m 9^ 0 va m # 1, phuong trinh vd nghiem. b) • Na'u X > 0, phuang trinh trd thanh : (2m - l)x = 1 - m m =— - phuang trinh vd nghiem. m 7^ — - phuang trinh cd nghiam x = 2 ' 2m-l
> 0 <:^ — < m < l . 2
• Na'u X < 0, phuang trinh trd thanh (2m + l)x = 1 - m m=—
phuang trinh vd nghidm.
<0 o m< hoac m >1. m i^ — , phuong tiinh cd nghiam x = 2 2m + l 2 Tdm Iqi: Vdi m > — hoac m < — Vdi 374
phuong trinh cd nghiem.
< m < — phuang trinh vd nghiem.
Cau 2. (2 d, mdi cau 1 diem) a) Ham sd ddng bien tran M vi he sd gdc a = 2 > 0. b) Do thi ham sd cit Ox tai x =^IL,
cit Oy tai y = m - 2. Khi dd dien
2
tich cua tam giac I a S = — ( m - 2 ) ^ = 6 < » m = 2±V6 4 cau 3. (1 diem). 5 1 x=---y=3 3
DesoS I.
CAU HOI TRAC NGHIEM K H A C H QUAN
Cau 1. Miln xac dinh cua ham sd y = x + Ixl +
, la v2-3x
(a) R.
(b) X < |
(c)x<^;
(d)x7^|.
Cau 2. Ham sd y = -2x + m - 1 (a) Ludn dong bien tran R. (b) Ludn nghich bien tran M. (c) Dong bia'n hoac nghich bien tan M tuy theo m. (d) Cd mdt gia tri ciia m de ham sd la ham sd hing.
375
1 1 cau 3. Phuang trinh -Jx-2+— = — 5 ^ 2 - x = 0X
2
(a) Cd nghiem la x = 1
(b) Cd nghiam la x = 1 va x = 2
(e) Cd nghiem la x = 2
(d) Vd nghiem.
Hay chgn kit qua diing. Cau 4. Phuang trinh 2x + 1 = 0 tuong duong vdi phuang trinh nao sau day? (b) I 2x1-1 = 0; (a) 4x^-1 =0; (c)4x^-4x+l =0; IL
(d)(x--).Vx^=0.
CAU HOI TULUAN
cau 1. Giai va bien luan phuong trinh sau: a) ( m ^ - l ) x + ( m - 1 ) = 0;
b) 3x = 2mx + m - 1.
Cau 2. Cho ham sd y = (m - l)x + 3x + m - 2. a) Xac dinh m de ham so dong biln tran M. b) Xac dinh m de do thi ham sd song song vdi dudng thing x + 3y = 1. cau 3. Cho phuang trinh x - 2 x + m - 3 = 0 . a) Xac dinh m de phuofng trinh cd hai nghiem trai da'u. b) Xac dinh m de phuong trinh cd hai nghiem ma nghiem nay ga'p ddi nghiem kia.
Hudng ddn vd ddp dn: I.
c A u HOI TRAC NGHlfiM K H A C H QUAN Cau 1 (1 d). Chgn (c). Cau 2 (1 d). Chgn (b). Cau 3 (1 d). Chgn (c). Cau 4 (1 d). Chgn (c).
376
n.
CAu HOI TULUAN cau 1. (2 d, mdi cau 1 d) a) Vdi m = 1, phuang trinh cd vd sd nghidm; Vdi m = - 1, phuang trinh vd nghiem. 3 b) Vdi m =— . phuang trinh vd nghiem 3 1—m Vdi m ^ —. phuang trinh cd nghiem duy nha't x =;2 m - 3 cau 2. (2 d, mdi cau 1 diem) a) Ham sd dong biln tran R khi m > - 2 . 1 7 b) m + 2 = — hay m = — 3
3
cau 3. (3 diem, mdi cau 1,5 d). a) m - 3 < 0 hay m < 3. b) A'= 4 - m > 0 hay m < 4. Khi dd 'Xi+X2=2 X1X2 = m - 3 Xi=2X2
35 Giai ha phuong trinh ta dugc m = — 9
De so" 4 L
CAU HOI TRAC NGHIEM K H A C H QUAN
Cau 1. Xet phuang trinh x + — = 1 (1) X
Phuang trinh nao sau day khdng tuong duong vdi phuong trinh (1)? (a)x^+Vx=-l;
(b)l2x+ll+ V2x + 1 = 0.
377
(c)xVx^=0;
(d)7+ V6x-1 = - 1 8 .
cau 2. Ham sd y = 7x + 13x1 + I 2 x + 171 (a) Ludn dong bien trdn M ;
(b) Ludn nghich bien trdn
(c) la ham sd hing;
(d) la ham sd bae nhit.
c a u 3. Tap gia tri eua ham sd y = 2x + V 5 - x
- A / X - 5 la :
(a) E.
(b) [5; + oo).
(c){5}.
(d) {10}. 2
cau 4. Khdng giai phuong trinh x +7x - 12 = 0. Hay diin diing-sai vao d trd'ng.
IL
(a) ludn cd hai nghiem; (b) ludn cd hai nghiem trai diu;
U Ll
(c) vd nghiem
U
(d) Ludn cd hai nghiem duong
U
CAU HOI TULUAN
cau 1. Cho ham sd y = x + 3x + 2. a) Xac dinh true dd'i xiing cua do thi ham sd. b) Cho diem M thudc do thi ham sd va cd hoanh do la 5. Hay xac dinh M' dd'i xiing vdi M qua true dd'i xiing ciia do thi. cau 2. Giai he phuang trinh sau f3x-4y = 6 -^ [x-3y = 4 [x-3y = 2 Cau 3. Cho phuang trinh 2x + yjx-l = m - 1. f-3x-2y = l
a)
b)
a) Giai phuong trinh khi m = 5. b) Xac dinh m de phuong trinh ed nghiem. 378
Hudng ddn vd ddp dn: L
CAU HOI TRAC NGHlfiM K H A C H QUAN cau 1 (1 d). Chgn (b). cau 2 (1 d). Chgn (a). cau 3 (1 d). Chgn d). cau 4 (1 d). (a) D
n.
(b) D
(c) S
(d)S
CAU HOI TULUAN cau 1. (2 d, mdi cau 1 d) 3 a) True dd'i xiing la x = — /
T\
b) y^= 42, => yM-= 42. Ta cd xj^+x^^^' =2. =-3, V ^J cau 2. (3 d, mdi cau 1 diem)
XM'
-3-5=-8.
5 13 a) x = — ,y = 11 11 b)x = 2,y = a. cau 3. (3 diem). Dat ylx-\
= t > 0. Khi dd ta cd f^ + r - m - 3 = 0
a) Vdi m = 5, ta dugc t^ =l,t2= -2 (loai). Vdi t = 1 ta cd X = 2. b) Phuang trinh da cho cd nghiem khi A>0 4m-ll>0 { + 3>0 hoac t i t 2 > 0 <=>-m ti+t2>0
(Diiu nay khdng xay ra).
-i>0 2
hoac ^1^2 ^ 0 bay -m + 3 < 0 hay m > 3.
379
Deso5 I.
CAU HOI TRAC NGHlfiM K H A C H QUAN
cau 1. Cho A = (1; 5), B = (m ; +oo). A n B la mdt khoang tran true so khi: (a)m
(b)l
(c) m < 5;
(d) m > 5.
Hay chgn kit qua sai. 2
cau 2. Ham sd y = -3x + 2x + m - 1. Hay chgn kit luan sai trong cac kit luan sau day: (a) Ludn cd gia tri Idn nhit. 2 (b) Hoanh dd dinh la x = — 3
(c) Ludn cit true hoanh. (d) Cit true tung tai y = m - 1. cau 3. Phuang trinh 4x-2+-
= 0: x
(a) Vd nghiam.
(b) Cd nghiam
(c) Cd hai nghiem
(d) Tap nghiam la tap xac dinh
cau 4. Cap phuang trinh nao sau day tuong duong. (b) (x^ + 1) x = 0 va Ixl = 0; (a) x(x - 1) = 0 va X ^ = 1; (c) x+ -j= = -j= va X = 1; yjX
IL
(d) x. V x - 3 =0 va 3x + 35 = 0.
yjx
CAU HOI TULUAN
cau 1. Tim cac gia tri cua m de cac phuang trinh sau chi cd mdt nghiam. a)mlxl + ( m ^ - l ) = 0; cau 2. Cho ham sd y = 42-x 380
b) m V x - 2 = 0. + 42 + x
a) Chiing minh ring ham sd tren la ham sd chin b) Xac dinh m de phuong trinh V2-X + v2 + x = m cd nghiem duy nhat. Cau 3. Giai he phuang trinh
1 - 1 =6 X
y
5 3 ^ — + — = -3
Hudng ddn vd ddp dn: I
CAU HOI TRAC N G H I £ M K H A C H QUAN cau 1 (1 d). Chgn (d). Cau 2 (1 d). Chgn (c). cau 3 (1 d). Chgn (a). cau 4 (1 d). Chgn (b).
n.
CAu HOI TIT LUAN cau 1. (2 d, mdi cau 1 d) a)m = ± 1. b) m > 0. Cau 2. (3 d, mdi cau 1,5 d) a) Dua vao dinh nghia ham sd chin. b) Phuang trinh cd nghiem duy nha't khi x = 0, do dd m = 2v2 2 2 cau 3. x = — y = — 3 3
381
De so 6 I.
CAu HOI TRAC NGHlfiM K H A C H QUAN
Cau 1. Cho A = (-00 ; 5), B = (m ; +oo). A n B la mdt khoang tran true sd khi (a) Vm e M;
(b) m = 6.
(c) m < 5;
(d) m > 5.
Hay chgn kit qua dung. 2
Cau 2. Ham sd'y = m x + m - 1 (a) Ludn dong bie'n trdn M. (b) Ludn nghich bien tran E. (c) Dong bien hoac nghich bien tan E tuy theo m. (d) Cd mdt gia tri ciia m de ham sd la ham so hing. cau 3. Phuang trinh 4x-2+
.
= 0 cd diiu kidn xac dinh la :
ylx-2 (a) E.
(b) [2; + oo).
(c)(2; + oo).
(d) E \ { 2}.
Cau 4. Phuong trinh Ixl = 4 tuong duong vdi phuang trinh nao sau day.
IL
( a ) x ( x - 2 ) = 0;
(b) (x^ + 1) Ix I = 4;
(c) (x + 2)(x - 2);
(d) x. y/x + 2 = 0.
CAu HOI TULUAN
cau 1. Giai cac phuong trinh sau: a)
+ x+1
382
x-2
=- - ; 2
b)- +- =5 X 2
cau 2. Cho ham sd y = (m - l)x^ - 3x + m - 2. a) Xac dinh m dl ham sd nghich bie'n trdn E. b) Vdi m ^ 1 hay xac dinh m de ham sd cd gia tri ldn nhit. c) Vdi m tim dugc d cau b), xac dinh gia tri Idn nha't cua ham sd. Cau 3. Cho phuong trinh 2
X + 2mx - m + 1 = 0 a) Xac dinh m de phuong trinh cd hai nghiem phan biet. b) Xac dinh m de phuong trinh cd hai nghiem ma hieu hai nghiem bing 1.
Hudng ddn vd ddp dn: L
c A u HOI TRAC NGHIEM K H A C H QUAN cau 1 (1 d). Chgn (c). cau 2 (1 d). Chgn (d). cau 3 (1 d). Chgn (c). cau 4(1 d). Chgn (c).
IL
CAu HOI TULUAN cau 1. (2 d, mdi cau 1 d) a) DK: X 7^ - 1 , X 7^ 2. Giai ra ta cd x = 1 va x = — b) X = 5 ± 421 cau 2. (3 d, mdi cau 1 diem) a) Ham so nghich bien tran E khi m = 1. b) Ham sd cd gia tri Idn nhat khi m - 1 < 0 hay m < 1. 2
N / - - . • 1^ w^u ' u--i-"-^ ~4m +12m + l c) Gia tri Ion nhat cua ham so l a — = 4a 4(m-l) 383
cau 3. (1 d) a) A'=m^ + m - 1 >Okhim< - l - V s h o a c m > - l + V5 b ) ( x 2 - x i ) =('xi-X2J -4xiX2 = m ^ + 4 m - 4 = lc:>m = l,m = - 6 . So sanh vdi diiu kien ta dugc m = 6.
D e SO 7 I. CAU HOI TRAC NGHIEM K H A C H QUAN cau 1. Diem nao trong cac diem sau day thudc dd thi ham sd y = x + 4x 7 (a)(-l;2);
(b) (1; 2);
(c)(-2;4);
(d) (2;
2-42)
Cau 2. Ham sd y = 21x1 + m - 1 (a) Ludn dong bien tran E. (b) Ludn nghich bien tran E. (c) Dong bien (0; + co) va nghich bien tran (-oo; 0][ (d) Cd mdt gia tri ciia m de ham sd la ham sd hing. Hay chgn kit qua diing. cau 3. Phuong trinh Vx+— = 0 cd diiu kian xac dinh la : X
(a) E;
(b) [0; + oo);
(c)(0;+co);
(d) E \ { 0 } .
cau 4. Phuong trinh x + Ixl = 2 tuong duong vdi phuang trinh nao sau day?
384
(a) (X- 1) = 0;
(b) (x^ + 1) X = 0;
(c)x+ -j= =^=; V^ vx
(d)x.4x-3
=0.
IL
CAU HOI TULUAN
Cau 1. Giai va bien luan phuong trinh sau: a) x^ - 2x + m^ + m - 1 = 0
b) mx^ = 2m x + m - 1 .
Cau 2. Cho ham sd y = (m - l)x + m - 2. a) Xac dinh m de ham sd dong biln tran E. b) Vdi m 7^ 1, hay xac dinh m de ham sd cit cac true toa do tao thanh mdt tam giac cd chu vi bing 6. cau 3. Cho phuong trinh x^-2x + k - 1 = 0 Xac dinh k de phuong trinh cd hai nghiem la nghich dao cua nhau.
Hudng ddn vd ddp dn: I.
CAU HOI TRAC NGHlfiM K H A C H QUAN cau 1 (1 d). Chgn (b). cau 2 (1 d). Chgn (c). cau 3(1 d). Chgn(c). Cau 4 (1 d). Chgn (a).
IL
CAU HOI TULUAN cau 1. (2 d, mdi cau 1 d)
a) A' = -(m-l)(m+2). Tir dd ta cd mG[l; 2], phuong trinh cd nghiem, eac trudng hgp con lai phuong trinh vd nghiem. 2
b) +) Ndu m = 0 phuong trinh vd nghiem; Neu m 7^ 0, A = 2m - m. Tir dd ta cd : me (-oo;0) u [— ; +00) phuong trinh cd nghiem. Trudng hgp edn lai phuang trinh vd nghiem. cau 2. (2 d, mdi cau 1 diem) 385 25-TKBGBAIs6lONC-T1
a) Ham sd dong bien tran E khi m >1.
b)S= 2
(m-2)^ m-1
= 6 khim = 8±V48
c a u 3. (2 diem). A' = 2 - k. De phuang trinh cd hai nghiem thi k < 2. Phuang trinh cd hai nghiem la nghich dao cua nhau khi X1X2 = 1 hay k - l = l , k = 2 (khdng thoa man).
De so 8 L
CAU HOI T R A C NGHlfiM K H A C H QUAN
c a u 1. Cho A = {-1 ;1; 2; 3};
B = {-1; 2; 5}.
Hay diin diing - sai vao cac cau sau: (a) A u B = {-1; 1; 2; 3; 5}
DDiing
DSai
(b) A n B = 0
DOung
nSai
(c)A\B{l;3}
DDiing
DSai
(d) (A u B) \ B = A u B
DDiing
DSai
c a u 2. Mdt hinh chii nhat cd chilu dai 3m ± 0,1m, chilu rdng 2m ± 0,2m, khi dd chu vi P bing (a) 5m ± 0,1m (b) 5m ± 0,2m (c) 5m ± 0,3m (d) 5m ± 0,6m Hay chgn kit qua diing. cau 3. Cho 2 dudng thing (d,) : x + y + m = 0; (dj) : 2x + my - 1 = 0 Hay diin diing - sai vao cac cau sau 386
(a)d,//d2C^m = 2
DDiing
DSai
(b)d, = d 2 < » m = - 2
DDiing
DSai
(c)di citd2 c:>m;^±2
DDiing
DSai
(d) dl cit d2 ce> m = ±2
DDiing
DSai
cau 4. Hay chgn kit qua diing trong cac kit qua sau (a) yjx-l = X
<=> X - 1 = x^
(b) V x - l = X
<=>x-l=x
(C) yjx-l
= X
(d) Vx - 1 = X IL
vax>l
<:::>X-1=X^+1
«>X = X^+1
CAu HOI TULUAN
Cau 1. Giai va bien luan phuang trinh sau: a) x-1
+— ^ =1 x-2
Cau 2. Cho ham sd y =
h) (x-2)(x^-2x
+ m) = 0
x-4
a) Chiing minh ring ham sd ludn dong bien tran tap xac dinh. b) Xac dinh toa do giao diem cua dd thi ham sd trdn vdi dudng thing y = 2x - 3. Cau 3. Giai va bien luan phuang trinh x - 2x + m = 0.
Hudng ddn vd ddp dn: L
CAU HOI TRAC NGHlfiM K H A C H QUAN Cau 1 (1 d). (a) D
(b) S
(c) D
(d) S
(b)S
(c) D
(d) S
cau 2 (1 d). Chgn (d). cau 3. (1 d) (a) D
387
Cau 4 (1 d). Chgn (b). IL
CAU HOI TULUAN Cau 1. (3d, mdi cau 1,5 d) a) Diiu kien phuong trinh : x i^ I va x i^ 2. Phuong trinh tuang duong vdi: x^ - 2(m - 2)x -2m + 4= 0. Ta cd A' = m - 2m. +) Vdi m G (0 ; 2) phuong trinh vd nghiem.
+) Vdi m = 0 hoac m = 2 phuang trinh cd nghiem kep (sau khi da so sanh nghiem vdi 1 va 2). +) Vdi m < 0 hoac m < 2, phuang trinh cd hai nghiem. Do x ^ 1 hoac X 7^ 2, nan thay 1 va 2 vao phuang trinh ta dugc m 7^ — va m 7^ — 4
6
Kit ludn : Ne'u m e (0 ; 2) phuong trinh vd nghiem 1 f\
Neu m = 0 hoac m = 2, m = — . phuong trinh cd 1 nghiem. 6 Neu m e (-00; 0) u(2; — ) u ( — ;+oo), phuang trinh cd hai nghiem phan biet. 6 6 b) Dat f(x) = x^ - 2x + m. Kit ludn: Neu m > 1, Phuong trinh cd nghiem x = 2. Neu m = 1, phuong trinh cd nghiem x = 1 va x = 2; Neu m = 0 phuong trinh cd hai nghiem x = 2 va x= 0 Neu m < 1, va m 7^ 0, phuang trinh cd 3 nghiem. Cau 2. (2 d, mdi cau 1 diem) a) Dua vao dinh nghia.
388
b) Hoanh dd giao diem la nghidm phuong trinh:
= 2x + 3 (Khdng cd x-4
giao diem). cau 3. (1 diem). Dat x^= t > 0, phuong trinh trd thanh : t^ -2t + m = 0
(2)
A' = 1 - m . Kit ludn: Na'u m = 1, phuang trinh (2) cd nghiem kep t = 1 d o d 6 x = ±l, Na'u m > 1, (2) vd nghiem, do dd phuong trinh da cho vd nghiem. Ne'u ac < 0 hay m < 0, phuong trinh (2) cd mdt nghiem duong, phuong trinh da cho cd hai nghiem Nlu 0 < m < 1, (2) ed hai nghiem duong, phuong trinh da cho cd 4 nghiem phan biet.
Deso9 L
CAU HOI TRAC NGHIEM K H A C H QUAN
Cau 1. Cho ham sd f(x) = Vx + 1 + V 5 - x xac dinh tran tap so nguyen. Khi dd tap xac dinh ciia ham so y = f(x) la ( a ) D = { 0 ; 1}
( b ) D = { 0 ; 1; 2}
(c) D = {0; 1; 2; 3; 4}
(d) D = {0; 1; 2; 3; 4; 5; -1}
Hay chgn kit qua diing. cau 2. Hay chgn cau la menh de trong eae cau sau day (a) Miia xuan hay cd mua phCin. (b) Nudc d Ho Guam ban qua. (c) 2758 la sd nguydn to. (d) ca ba cau trdn khdng la menh dl.
389
Cau 3. Cho ham sd f(x) = (x - l)(x - 2) + x + 3. (a) Ham sd dong bien tran E (b) Ham sd nghich bien tran E (c) Ham sd dong biln trdn (5; + oo) va nghich biln tran (-oo; 5) (d) Ham sd la ham hing. Hay chgn kit qua dung. cau 4. Tap xac dinh ciia ham so f(x) = V x - l + Vx + 2 x - 5 la (a)x>l (c)x>l+V6 IL
(b)x<-l-V6 ( d ) l < x < l + V6
cAU HOI TULUAN
cau 1. Giai cac he phuong trinh {2x-y = 5 a) '[x-6y = 3
b)
[3x + 2y = 4 -^ [2x + 4y = 5
2
cau 2. Cho ham sd y = x + 3x +2. a) Xac dinh giao diem ciia do thi ham sd vdi Ox va Oy. b) Tinh chu vi tam giac cd dinh la ba diem d cau a). c) Tfnh dien tfch tam giac cd dinh la ba dilm d cau a).
Hudng ddn vd ddp dn: L
CAU HOI T R A C NGHIEM K H A C H QUAN cau 1 (1 d). Chgn (d). cau 2 (1 d). Chgn (c). cau 3 (1 d). Chgn (c). cau 4 (1 d). Chgn (c).
390
n.
CAU HOI TULUAN cau 1. (3 d, mdi cau 1,5 d) '^_27 11 a) " 1
b).
x=— 4 7 ^=8
Cau 2. (3 d, mdi cau 1 d) a)(0;2),(-l;0),(-2;0).
b) P=2V2 + 1 + V5 c)S=l
391
De so" 10 I. CAU HOI TRAC NGHIEM K H A C H QUAN Cau 1. Cho 2 tap hgp A va B cd A n B ?t 0
Kf hieu | M | la sd phin tii ciia tap M.
Khidd (a) | A u B | < |A| + | 5 |
(b)|AuB|>|A|
(e) | A u B | > | e |
(d) | A u B | = |A| + |5|
Hay chgn ket luan sai trong cac kit luan tran. Cau 2. Cho hai menh dl P : 7 la hgp so" : 2
- 1 la so nguyen to
Khi dd menh dl P =i> Q (a) diing
(b) sai
(c) khdng la menh dl
(d) ca ba kit luan trdn diu sai.
cau 3. Cho f la ham sd chin, g la ham so le Khi dd S(x) = f(x) + g(x) (a) la ham sd chin
(b) la ham sd le
(c) la ham sd khdng chan, khdng le
(d) ca ba kit luan tran diu dung.
cau 4. Cho f va g nhu cau 3. Khi dd S(x) = f(x).g(x) (a) ham sd chin
(b) ham sd le
(c) ham sd khdng chan, khdng le
(d) ca ba kit luan diu sai.
Cau 5. Cho d : y = 2x + m; p : y = X + 2mx + 1. Dieu kien de a va P cit nhau tai 2 diem phan biet la
392
(a) 0 < m < 1
(b) m < 0 hoac m > 1
(e) 0 < m < 1
(d) m < 0 hoac m > 1
IL
CAU HOI TULUAN
cau 1. Giai va bien luan phuang trinh sau: a) (m + l)x + (m^ -1) = 0;
b) 2x^ = 2mx + m - 1.
cau 2. Cho ham sd y = (m^ - m - l)x^ + 2x + m - 2. a) Xac dinh m de ham so dong bien tran E. b) Hay xac dinh m de ham sd cd gia tri ldn nhit. Cau 3. Giai phuong trinh a) 2x + V x - 2 +2 = 6; b)Vx-l=|x-2|.
Hudng ddn vd ddp dn: I.
CAU HOI TRAC NGHlfiM K H A C H QUAN cau 1 (1 d). Chgn (d). cau 2 (1 d). Chgn (a). cau 3 (1 d). Chgn (c). Cau 4 (1 d). Chgn (b). cau 5 (1 d). Chgn (b).
IL
CAU HOI TULUAN cau 1. (2 d, mdi cau 1 d) a) Vdi m = - 1 , phuang trinh cd vd so nghiem; Vdi m = 1, phuong trinh vd nghiem. Vdi m 7^ ± 1, phuang trinh cd nghiem duy nha't b)A = m - m + l > 0 , phuong trinh ludn cd hai nghiam phan biet cau 2. (2d, mdi cau 1 d) a) m < 1- Vs hoac m > l+Vs 393
26-TKBGDAISO10NC-T1
-'-'-'
b) 1 - V 5 < m < l + V5 cau 3. (2d, mdi cau 1 diem) a) Dat V x - 2 = / > 0. Phuang trinh tra thanh 2t^ + t = 0 ta dugc t = 0 hay x = 2. 5±V5 b) Binh phuang hai vd ta cd : x =
De so 11 I.
CAU HOI TRAC NGHlfiM K H A C H QUAN
Cau 1. Cho A va B la hai tap A n B ?t 0 . Kf hieu | Q | la sd phin tii cua tap Q. Khi dd: (a)|A\5|<|A|
(b)|A\B|<|5|
(c)|5\A|<|5|
(d) | 5 \ A | = | A u B | - |A|
Hay chgn kit luan sai. cau 2. Cho A = {x € Rl X - 2 > 0}; B = {x e Rl x^ + 2x - 3 < 0}. Khi dd A n B la: (a){xeRlx>2}
(b) {x e R l - 3 < x < 0}
(c) 0
(d) mdt kit qua khac.
Cau 3. Ham sd : y = 3x + | x - l | +|x + l[. (a) ludn ludn ddng bien
(b) ludn ludn nghich bien
(c) la ham sd hing
(d) mdt kit qua khac.
cau 4. Phuang trinh | x - l | = 2 cd sd nghiam la (a)-l
(b)2
Hay chgn kit qua sai. 394
(c)3
(d){-l;2}.
IL
C Au HOI TRAC NGHI$M TU LUAN
cau 1. Giai va bien luan he phuong trinh sau: \ax + y = 2
fx + y = 2
[x-2ay = 3
[ax-2y = l
Cau 2. Cho ham sd y = (2m - l ) x + 4x + m - 2 . a) Xac dinh m de ham sd dong bie'n trdn E. b) Hay xac dinh m de ham sd cd gia ldn nha't. cau 3. Giai phuang trinh |x-l| +|x-2| = 3 Hudng ddn vd ddp dn: I.
CAU HOI TRAC NGHIEM K H A C H QUAN cau 1 (1 d). Chgn (b). Cau 2 (1 d). Chgn (c). cau 3 (Id). Chgn (a). cau 4 (I d). Chgn (b).
IL
CAu HOI TRAC NGHIEM TU LU^N cau 1, (2 d, mdi cau 1 d) a) D = -2a -I i^O, phuang trinh cd nghiem duy nhit. b) D = - 2 - a. Nlu a = - 2 phuang trinh vd nghiem Neu a ^ - 2 phuong trinh cd nghiem duy nha't. cau 2. (2 d, mdi cau 1 dilm) 1 a) m >—; 2 1 b)m < — 2 cau 3. (2diem). x = 0 va x = 6. 395
PHU LUC
Ngudi ta bit diu phai nghi den doi mdi phuang phap day hgc d Viet Nam trong thdi dai vi tfnh hoa. cau hdi dat ra la ta phai bit diu tir dau? Chic chin cau tra Idi cua nd la: Phai nghian ciiu va phai hgc. Ta phai nghian ciiu de iing dung nhiing cai ma the gidi da cd vao thuc tiln Viet Nam. Ta phai hgc vi neu khdng ta ching lam gi dugc. Ddi vdi mdt giao vidn trong thdi hien dai nay viec diu tian la phai co trinh do tin hgc co ban. Dd'i vdi giao vian Toan diiu dd lai cin thiet va cap bach ban rat nhilu. Sau day tdi xin gidi thieu mdt vai iing dung phin mim Maple trong qua trinh giang day dai so 10. A.
GlOl THIEU \ t PHAN MEM MAPLE
Phang mIm nay cho ta tdc do tfnh toan cue ki td't, md phdng hinh hgc mdt each true quan, giai phuong trinh va he phuang trinh thuan tien va mdt sd ling dung khac nira. Dac biet la phan mIm nay cd giao dien vdi window ra't td't. De gidi thieu vl nd cac ban, dgc tren cac webside ciia cac diin dan toan hgc. B. MOT SO tlNG DUNG vA VI DU I.
Tinh toan va tinh toan vdi cac sd gan dung Sau khi vao Maple, ta tfnh cac gia tri cua cac bieu thiic so vdi cac kf hieu
sau: Phep cdng :
Dau +;
Phep nhdn :
Dau *;
Phep trie :
Dau ;
Phep chia :
Dau /;
Phep ndng len luy thita : 396
Dau ^
Phep tinh lay can bdc hai: sqrt Ngoai ra ta ciing dung cac dau ngoac (, {, ), },[,],••• Vi du 1. T' K linh
1999^-2007'
^ ,. , . ta lam nhu sau :
2005^+2008.2001 > (1999^2-2007^2)/(2005^3+2008*2001); -32048 8064168133 Sau khi nhap xong cac sd lieu an "Enter" may se cho ta kit qua.
Vi du 2: Tfnh gia tri ciia bieu thiic chinh xac din mudi chii sd thap phan
Vi7+Vi3-2Vn 6V23 Ta lam nhu sau:
> (sqrt(17)+sqrt(13)-sqrt(ll))/(6*sqrt(23)); — (Vi7+Vl3-Vn)V23 138
> e v a l f (%) ; .1533287151
Chii y rang lenh > e v a l f (%) ; Cd nghia la tfnh gia tri cua bieu thiic ngay trudc dd de'n 10 chii sd thap phan. Nlu ta cin tfnh ngay trudc dd den sd chir sd thap phan bat ki, ta cd the lam nhu sau de tfnh gia tri bieu thiic tren din 10 chir sd (sau khi da nhap xong bieu thiic) >evalf((sqrt(17)+sqrt(13)-qrt(ll))/(6*sqrt(23)),5); .15333 Nhu vay neu trong chuang I, viec tfnh toan vdi sd gin diing se vd ciing thuan Igi nlu ta sii dung phin mIm nay. 397
Vidu 3: 17 99 Trong hai sd— va — sd nao gin V2 ban (Bai tap dai sd 10 nang cao) 17 70 12 Tuy vao dau ciia bilu. thiic nay ma ta danh gia kit qua. Viec nay cd the sir dung may tinh bd tiii, song khdng dan gian vi nd cd gia tri tuyet dd'i.. De giai bai toan nay la chi cin thuc hien phep tinh
V^-
Ta thuc hien vdi Maple nhu sau: > (abs(sqrt(2)-17/12))-(abs(sqrt(2)-99/70)) ; 1 420 99 . r Ta tha'y gia tri nay duong, do do — gin v2 ban. Vidu 4: Hay danh gia sai sd tuyet dd'i ciia so Pi ( TI ) vdi so
355 113
Gidi: Ta cd sai sd tuyet dd'i A =
%
-
•
355 113
Sii dung phin mIm Maple bing cii phap sau ta cd: >abs(Pi-355/113) ; -Tt +
355 113
>evalf(abs(Pi-355/113),8); .2 10"^ Nhu vay, neu lay chinh xac den 8 chii sd thap phan thi ta tha'y A < 0,2.10"^
398
Ta cd the giai rat nhilu vf du, (trong sach giao khoa, sach bai tap 10) bing Maple ra't td't. II.
Ham sd va gia tri ciia ham sd, do thi cua ham so Ta hoan toan cd the tfnh dugc tfnh dugc gia tri ciia ham sd tai cac bien sd.
/.
Ddu tien chung ta dinh nghia hdm so, chdng hqn hdmf(x) = x^ + — > f:=x->x^2+x+l/3; /• 2 1 /:=x^x+x +— > f (1) ; 7_ 3 >f (sqrt(2)) ;
1 + 42 3
>f (l-sqrt(3) ) ;
(1-V3)2+--V3 >evalf(%); .137180910 > f (Pi) ; 1
2 Tt
+Tt+-
>evalf(%); 13.34453039 > f (1/2) ; \3_ 12 399
>f (sqrt(3)-sqrt(2)) ; (V3-V2)2+V3-V2+-!3 >evalf(%); .7521910943
2.
Ngudi ta cd the dung lenh unapply de chuyen mdt bieu thiJCc ve mot hdm sd, chdng hqn > g : = u n a p p l y ( x ' ^ 2 + s q r t (2) * x + l - s q r t (2) ,x) ; g := X -> x^ + V2x +1 - V2
va tir dd ngudi ta cd the xac dinh dugc cac diem cin thiet ciia dd thi ham so, chang ban 7
•).
v 2
Gia tri cue tieu: Tai x =
khi do gia tri cue tiau la 2
6
. .
>g(-sqrt(2)/2) ;
i-V5 2
> evalf(%) ; .9142135620 Ta con xac dinh dugc gia tri xap xi ciia nd Giao diem vdi true tung >g(0) ;
I-V2 > e v a l f (%) ; 414213562 3.
Hdm tdng khuc De xay dung ham tu"ng khiic ta diing lenh: Piecewise.
400
Ching han
>f :=piecewise(x<=-l,x'^2+l,x<=l,-abs (x)+1, s i n (x+1) /x) /
f:=
x^+1
x<-l
-|x| + l
X<1
sin(x +1)
otherwise
> f (0)
x^+1
x<-l
-|x| + l
X<1
sin(x + l)
(0)
otherwise
>evalf(%)
x^+1
x<-l
-|x| + l
X<1
sin(x + l)
4.
(0)
otherwise
Do thi hdm sd (trong mat phdng) De ve do thi ham sd dau tidn ta lam sach bd nhd bing lenh: Restart
Sau dd nap gdi lenh chiic nang chuyan dung de ve do thi bing hai lenh lian tia'p: With(plots) With (plotttools). > restart; >with(plots) ; Warning, the name changecoords has been redefined 401
[animate^ animate 3 danimatecurv^ arrow, changecoord?complexplofcomplexplot3(!l conformal conformaUdcontourplol contourplot3dcoordplo( coordplot3d cyliriderplo(densityplo(di.<;playdisplay34fieldplot,fieldplot3cigradploi gradplol3dimplicitplot implicilpIot34inequal listcontplol Ustconlplot34 listdensityploflistplot, listplot3d loglogplo( logplot matrixplot odeplotparetq pointplolpointplot34polarplot,polygonploipolygonplui3dpolyhedra_supporte(, polyhedraplofreplot, rootlocu^ semilogplotsetoption^ setoptions34spacecurvq sparsematrixplgtsphereplofsurfdatq textplot, textplot34 tiibeplot] >with(plottools); Warning, the name arrow has been redefined
[arc, arrow, circle, cone, cuboi4 curve, cut in, cutout, cylinder, disk, dodecahedron ellipse, ellipticArc, hemisphere hexahedrori homothety hyperbolq icosahedror; line, oclahedroripieslice, point, polygon, project, rectangle, reflect, rotate, scale, semitorus, sphere, stellate, tetrahedron torus, transform, translate, vrml] Sau dd ta ve do thi ham so bing cac cii phap sau: Vedd thi hdm thdng thudng
,.
Ta ve do thi ham sd y = f(x) bing cii phap Plot(f(x),x=a..b,y=c..d,titie= 'abed'); Ching ban Ve dd thi ham sd y = 2x
Vl3
>plot(2*x-sqrt(13),
402
x=-10
10);
Vdi do thi ham sd trdn, ta khdng dua c, d vao thi chuong trinh tu ddng xac dinh miln gia tri y. > p l o t ( 2 * x ^ 2 - 3 * x + 5 , x = - 2 . 4,y=2 . 1 0 ) ;
10 9 8i
\i y
\6 \.
I—'—'—'—'—r
-2
-1
—I—1—1—1—1—I—1—I—I—1—I
'^ 0
1
2
3
4
Ta cd the ve hai do thi ciia hai ham so trdn ciing mdt he true
>plot([2*x^2-3*x+5,2*x+l],x=-2..4,Y=2..10,color=[red,blue]);
403
Su van ddng cua dd thi Khi ham so cd chiia tham so, thi do thi cua nd thay doi theo tham sd. Khi tham sd biln thidn thi do thi thay doi, ta ndi do thi bien thian theo tham sd. De lam, viec diu tian ta ciing nhap chuang trinh. Sau dd, ta lam sach bd nhd bing lenh: Restart Sau dd nap gdi lenh chiic nang chuyan dung de ve do thi bing hai lenh lian tia'p: With(plots)
With (plotttools), nhu tran. Ching ban
> animate ( x ' ' 2 + 3 * t * x - t - l , x = - 5
404
. 5 , t = - l . . 1 , color=red) ;
40-
30-
20-
10-
•
•
1'
•
-4
'—
0.
-2
4 X
Sau dd, ta danh dau vao do thi, in chudt ban trai, vao nhirng not diiu khien ma ta thfch, sau dd a'n Play. IV. Giai phuang trinh va he phuang trinh Budc 1. Tao lap phuang trinh bing each gan cho phuang trinh dd mdt kf hieu nao dd, chang han Eqn: =, PTl :=,... Budc 2. Dung lanh solve de giai phuong trinh Solve(eqn); Vi du > q n : = x'^3+2*x-3=0; eqn :=x^ + 2 x - 3 = 0 > solve(eqn) ; 1 1
1 1
2
2
i,_i+l/Vn,-^-^/Vn 2
2
Ta tha'y phuong trinh cd nghiem thuc x = 1 va hai nghiem phiic
- - - - / V n va - - + - / V n 2
2
2
2
405
Vi du >PT:=sqrt(x-1)=3*x-5; PT:=4^
= 3x-5
>solve(PT) ;
Vi du >ab:= sqrt (5-x)+sqrt (x+3) =3; ab:=45-x +4x + 3 =3 > solve(ab) ;
l+lV7. 1-^V7 2
2
Vi du >solve((abs(x+abs(x+2))^2-1)^2=9.{x});
{x = 0},{x<-2} V.
Giai he phuang trinh
Budc 1. Dinh nghia cac phuong trinh cua he, gid'ng nhu dinh nghia trong giai phuang trinh. >solve((abs(x+abs(x+2))^2-1)^2=9,{x}); {x = 0}. {x<-2} > e q n l : = 2*x+3*y-4*z=12; eqnl •.= 2x + 3y-4z=
12
>eqn2:= 3*x-3*y-5*z=19; eqn2 : = 3 x - 3 > ' - 5 z = 19 >eqn3:= x+7*y-4*z=22; eqn3 :=x + 7y - 4 z = 22 Budc 2. Diing lenh solve dl giai he > solve({eqnl,eqn2,eqn3},{x,y,z}) ; f -334 -36 -2511 19 406
19
19
MUC LUC •
•
Trang Chuang L MENH Di - TAP HOP §1. Menh dl va menh dl chiia bia'n §2. Ap dung menh dl vao suy luan toan hoc Luyen tap §3. Tap hgp va eac phep toan tren tap hgp Luyen tap §4. So gin diing va sai sd cau hdi va bai tap dn tap chuong I Chuang n.
HAM SO BAC NHAT VA BAC
HAI
§1. Dai cuong vl ham sd Luyen tap §2. Ham sd bae nhit Luyen tap §3. Ham sd bae hai Luyen tap Cau hdi va bai tap dn tap chuong II Chuang ///.
PHUONG TRJNH VA HE PHL/ONG TRiNH
§ 1. Dai cuong vl phuong trinh §2. Phuong trinh bae nhat va bae hai mdt in Luyen tap §3. Mdt so phuang trinh quy vl phuong trinh bae nhat hoac bae hai Luyen tap §4. He phuong trinh bae nhat nhilu in Luyen tap §5. Mdt so vf du vl he phuang trinh bae hai hai in Cau hdi va bai tap dn tap chuang III ON
TAP HOC Ki I
PHU LUC
5 9 32 45 53 75 90 108 124 125 151 163 176 187 198 212 235
237 258 283 291 300 310 330 337 346 369 396
407
Chiu trdch nhiem xudt bdn : Giam ddc: DINH NGOC BAO Tong bian tap: LE A Chiu trdch nhiem ndi dung vd bdn quyen
CONG TY TNHH SACK GIAO DUC HAI ANH
Biin tap ndi dung : NGUVfeN TIEN TRUNG My thuat - Vi tinh : THAI SON SON LAM Trinh bdy bia : THANH HUYEN
THIJ^T KE BAI GIANG DAI SO 10 - NANG CAO, T^P MOT In 1000 cudn, kho 17 x 24cm, tai Cdng ti co phin in Phiic Yan. Sd dang kf KHXB : 219.2006/CXB/80 - 25/DHSP ngay 28/3/06. In xong va nop luu chieu thang 11 nam 2006.
