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ﺘﺼﺤﻴﺢ اﻹﻤﺘﺤﺎن اﻝﺠﻬوي 2015
اﻝدورة :ﻴوﻨﻴو 2015
اﻝرﻴﺎﻀﻴﺎت
اﻝﻤﺴﺘوى :اﻝﺜﺎﻝﺜﺔ إﻋدادي
اﻝﺘﻤرﻴن اﻷول -a-1ﻝﻨﺤل اﻝﻤﻌﺎدﻝﺔ 5 x − 3 = x + 9 : 5x − x = 9 + 3 4 x = 12 12 =x =3 4 ﺤل اﻝﻤﻌﺎدﻝﺔ ﻫو اﻝﻌدد 3
−2 x − 4 y + 2 x − 5 y = −6 − 12 −9 y = −18 −18 =y =2 −9 ﻨﻌوض yب 2ﻓﻲ اﻝﻤﻌﺎدﻝﺔ ): ( 1 + 2×2 = 3 +4=3 = 3−4 = −1
-bﻝﻨﺤل اﻝﻤﻌﺎدﻝﺔ : ﺤل اﻝﻨظﻤﺔ ﻫو اﻝزوج
( x − 4) × ( 3 x − 5) = 0 3 x − 5 = 0أو x − 4 = 0 3 x = 5أو x = 4 5 = xأو x = 4 3 5 اﻝﻤﻌﺎدﻝﺔ ﺘﻘﺒل ﺤﻠﻴن ﻫﻤﺎ 4 :و 3 x ≥ −2 x + 9 -2ﻝﻨﺤل اﻝﻤﺘراﺠﺤﺔ : x + 2x ≥ 9 3x ≥ 9 9 ≥x 3 x≥3 ﺤﻠول اﻝﻤﺘراﺠﺤﺔ ﻫﻲ اﻷﻋداد اﻷ ﻜﺒر ﻤن أو
) ( −1 ; 2
اﻝﺘﻤرﻴن اﻝﺜﺎﻨﻲ -1
120 + 340 + 100 + p = 600 560 + p = 600 p = 600 − 560 = 40 -2ﻝدﻴﻨﺎ أﻜﺒر ﺤﺼﻴص ﻫو 340وﻗﻴﻤﺔ اﻝﻤﻴزة اﻝﻤواﻓﻘﺔ ﻝﻬذا اﻝﺤﺼﻴص
ﻫﻲ 2 إذن اﻝﻤﻨوال ﻫو 2 120 -3اﻝﻨﺴﺒﺔ اﻝﻤﺌوﻴﺔ اﻝﻤواﻓﻘﺔ ﻝﻠﻤﻴزة × 100 = 20% : 0 600 -4اﻝﻤﻌدل اﻝﺤﺴﺎﺒﻲ : 0 × 1 20 + 2 × 3 40 + 3 × 10 0 + 4 × 4 0 60 0 0 + 68 0 + 3 0 0 + 16 0 = m = 1, 9 6 00 = m
ﺘﺴﺎوي 3 -a-3
2 × 1 − 5 × 1 = 2 − 5 = −3 ≠ −12 إذن اﻝزوج ) ( 1 ; 1ﻝﻴس ﺤل ﻝﻠﻨظﻤﺔ
x x x x
اﻝﺘﻤرﻴن اﻝﺜﺎﻝث -1
f ( x) = 4 x − 4
–aﺼورة اﻝﻌدد : 5
-bﻝﻨﺤل ﺠﺒرﻴﺎ اﻝﻨظﻤﺔ : x + 2 y = 3 )(1 ) 2 x − 5 y = −12 ( 2 ﻨﻀرب طرﻓﻲ اﻝﻤﻌﺎدﻝﺔ ) ( 1ﻓﻲ −2وطرﻓﻲ اﻝﻤﻌﺎدﻝﺔ ) ( 2ﻓﻲ : 1
−2 x − 4 y = −6 2 x − 5 y = −12 ﻨﺠﻤﻊ اﻝﻤﻌﺎدﻝﺘﻴن اﻝﻤﺤﺼل ﻋﻠﻴﻬﻤﺎ طرف ﺒطرف :
f (5) = 4 × 5 − 4 = 20 − 4 = 16
ﺘﺤدﻴد اﻝﻌدد اﻝذي ﺼورﺘﻪ : −8f (x) = −8 4x − 4 = 8 4x = 8 + 4 4 x = 12 12 = x = 3 4
اﻝﻌدد اﻝذي ﺼورﺘﻪ −8ﻫو 3
https://sites.google.com/site/stitmath
-c
-b
y = mx + p
f (2) = 4 × 2 − 4 = 8 − 4 = 4 ≠ 3
إذن اﻝﻨﻘطﺔ ) A ( 2;3ﻻ ﺘﻨﺘﻤﻲ اﻝﻰ اﻝﺘﻤﺜﻴل
ﺘﺤدﻴد : m
اﻝﻤﺒﻴﺎﻨﻲ ﻝﻠداﻝﺔ f
ﻝدﻴﻨﺎ
-a -2
إذن
3 x 4 -bﺘﺤدﻴد اﻝﻌدد اﻝذي ﺼورﺘﻪ : −6 g( x ) = −6
1 وﻤﻨﻪ 2 ﺘﺤدﻴد : p
= )g( x
) ( D ) ⊥ ( D2 m × ( −2 ) = −1 =m
اﻝﻨﻘطﺔ
)
A ( 2 ; 3ﺘﻨﺘﻤﻲ اﻝﻰ اﻝﻤﺴﺘﻘﻴم
3 x = −6 4 3 × 4 x = −6 × 4 4 3 x = −24 −24 = −8 3 اﻝﻌدد اﻝذي ﺼورﺘﻪ −6ﻫو −8
-a -1
) ( D2
1 y= x+ p 2 1 3= × 2 + p 2 3 = 1+ P p = 3−1 = 2
= x 1 و ﺒﺎﻝﺘﺎﻝﻲ x + 2 : 2 -a -2
اﻝﺘﻤرﻴن اﻝراﺒﻊ y = −2 x − 1
( D2 ) :
=y
( D2 ) :
( D) :
−2 × 2 − 1 = −4 − 1 = −5 ≠ 3 إذن اﻝﻨﻘطﺔ ) A ( 2 ; 3ﻻﺘﻨﺘﻤﻲ اﻝﻰ ) ( D -b y = mx + p
( D1 ) :
ﺘﺤدﻴد : m ﻝدﻴﻨﺎ
) ( D1 ) / / ( D
إذن m = −2 ﺘﺤدﻴد : p اﻝﻨﻘطﺔ
)
A ( 2 ; 3ﺘﻨﺘﻤﻲ اﻝﻰ اﻝﻤﺴﺘﻘﻴم
) ( D1 y = −2 x + p 3 = −2 × 2 + p 3 = −4 + p p = 3+4 = 7 و ﺒﺎﻝﺘﺎﻝﻲ y = −2 x + 7 :
( D1 ) :
uuur -bﺘﺤدﻴد : AB
) uuur
)
−4;−2
( ) = AB
A( 2 ; 3
) B ( − 2 ;1 uuur AB ( − 2 − 2 ; 1 − 3
-ﺤﺴﺎب : AB
https://sites.google.com/site/stitmath
-3 S × BF VBEFG = EFG 3 EF × FG 3 × 4 S EFG = = = 6m 2 2 2 6 ×1 VBEFG = = 2m 3 3
AB =
( −2 − 2 )
AB =
( −4 )
+ (1 − 3 )
+ ( −2 )
2
2
AB = 16 + 4 AB = 20 = 4 × 5 = 2 5
ﺒﺎﻹ زاﺤﺔ اﻝﺘﻲI ﻫﻲ ﺼورةI ' ﻝدﻴﻨﺎ-a-3
B إﻝﻰA ﺘﺤول uur uuur II ' = AB : إذن uur : II ' ﺘﺤدﻴد
-4 : ﻨﺒدأ ﺒﺤﺴﺎب ﺤﺠم اﻷﺴطواﻨﺔ V ' = s × h = 0,1 × 1 = 0,1m VABCDEFGH 12 = = 120 V' 0,1
2
2
3
120 ﻴﺘم اﺴﺘﻌﻤﺎل اﻝﺒرﻤﻴل
I ( 1; 0
)
I ' ( x I ' ; yI ' ) uur II ' ( x I ' − 1 ; yI ' − 0 ) uur uuur II ' ( x I ' − 1 ; y I ' − 0 ) = AB ( − 4 ; − 2
x I ' − 1 = −4
)
y I ' = −2
x I ' = −4 + 1 x I ' = −3 I '( − 3 ; − 2
)
: وﺒﺎﻝﺘﺎﻝﻲ
a -2 أﻨظر اﻝﻤﺒﻴﺎن ﻓﻲ اﻝﺴؤال-b
اﻝﺘﻤرﻴن اﻝﺨﺎﻤس B ﻗﺎﺌم اﻝزاوﻴﺔ ﻓﻲABC ﻝدﻴﻨﺎ اﻝﻤﺜﻠث-1 : ﺤﺴب ﻤﺒرﻫﻨﺔ ﻓﻴﺜﺎﻏورس اﻝﻤﺒﺎﺸرة إذن AC 2 = AB 2 + BC 2 : اﻝﺘطﺒﻴق اﻝﻌددي AC 2 = 3 2 + 42 AC 2 = 9 + 16 = 25 AC = 25 = 5m
-2 VABCDEFGH = S EFGH × AE VABCDEFGH = 4 × 3 × 1 = 12m 3
https://sites.google.com/site/stitmath
sujet(2015)+solution.pdf
تصØÙØ Ø§Ùإ٠تØا٠اÙجÙÙÙ 2015. اÙرÙاضÙات. اÙدÙرة : ÙÙÙÙÙ 2015. اÙ٠ستÙÙ : اÙثاÙثة إعدادÙ. Page 3 of 5. sujet(2015)+solution.
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