اأقدم لكم أسئلة رياضيات الفصل الثاني المثيرة لكن لكن لكن لكن لا تنسوني من الردود
لسؤال الأول /
إذا كانت أ = { 1 , 2 , 3 , 4 , 5 } , وكانت العلاقتين ع , م ممثلتين كالآتي :-
ع = { ( س , ص ) أ × أ , س > ص } .
م = {( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( 4 , 4 ) , ( 5 , 5 )} .
أ - مثل العلاقتين بيانيا : -
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ب - أوجد مجال ومدى كلا العلاقتين السابقتين .
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ج - بين أي العلاقتين علاقة تكافؤ ( مع ذكر السبب ) .
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السؤال الثاني /
أ ) بين أي من الاقترانات الآتية اقتران تناظر ( مع ذكر السبب ) :-
1 - ق ( س ) = س 3 .
2 - ك ( س ) = { ( 1 , 2 ) , ( 2 , 5 ) , ( 7 , 9 ) , ( 3 , 5 ) } .
3 - ﻫ ( س ) : ط ط
س 3 س – 2
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ثم : -
1- أوجد : ق ( 2 ) , ق ( – 2 ) , ك ( 7 ) , ﻫ ( 5 ) .
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2- إرسم المخطط السهمي للاقتران ك ( س ) .
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ص
ب ) هل العلاقة الموضحة بالرسم تمثل إقتران ؟ ولماذا ؟
وإذا كانت إقتران , هل هو إقتران واحد لواحد ؟ ولماذا ؟ س
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السؤال الثالث / إذا كان : -
ق ( س ) = 4 س – 8
ﻫ ( س ) = 2 س 4 + 3 س 2
ل ( س ) = س
1- بإستخدام الاقتران المحايد , أوجد الاقتران العكسي للاقتران ق ( س ) .
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2- ثم أوجد قيمة كلا من :-
ق 5 ق – 1 ( س ) : -
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ق – 1 5 ق ( 2 ) : -
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ﻫ 5 ل ( 4 ) : -
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السؤال الرابع /
إذا كان جذرا المعادلة س 2 + 2 س = 5 هما م , ن
كون المعادلة التي جذراها 2 م , 2 ن .
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السؤال الخامس /
إذ كان 2 س 2 – 6 س + 4 = أ ( س – م ) 2 + ن
أوجد قيمة كلا من أ , م , ن , ثم حل المعادلة بيانيا .
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السؤال السادس /
إذا كان طول مستطيل = ( س – 2 ) , و عرضه = ( س – 3 ) , وكان
قطر المستطيل = 5 سم , أوجد كلا من طول المستطيل وعرضه .
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السؤال السابع /
أوجد قيمة ( م ) التي تجعل الاقتران ق ( س ) = 3 – م س + س 2
يمس محور السينات .
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السؤال الثامن /
إذا كان : -
ق ( س ) = س 2 – 3 س + 2 .
ل ( س ) = س + 2 .
1- أوجد ناتج كلا مما يلي : - ( مع إيجاد درجة الاقتران الناتج ) .
أ- ق ( س ) – ل ( س )
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ب - ق ( س ) × ل ( س )
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ج - ق ( س ) ÷ ل ( س ) ( باستخدام القسمة المطولة ) .
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2 - بين هل الإقتران ل ( س ) عامل من عوامل الاقتران ق ( س ) .
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السؤال التاسع /
أوجد قيمة أ , ب التي تجعل أ س 3 – 3 س 2 – ب يقبل القسمة على ( س – 2 )
, ( س + 1 ) , ثم حلل كثير الحدود الناتج .
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السؤال العاشر /
حلل إلى العوامل الأولية كثير الحدود س 3 – 2 س 2 – 5 س + 6 .
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السؤال الحادي عشر /
إذا كان : -
ق ( س ) = س 2 – س – 6 , ل ( س ) = س + 2
س 2 – 6 س + 5 س – 1
أ – جد ناتج كلا مما يلي : ق ( س ) – ل ( س ) .
ق ( س ) ÷ ل ( س ) .
( ضع الناتج في أبسط صورة )
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ب – أوجد المجال لكل ناتج من الفقرة ( أ ) :
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السؤال الثاني عشر /
حل المعادلات الآتية :-
أ - 4 5 س + 3 ÷ 4 3 س + 1 = 32
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ب - 3 - س = 1 \ 81
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ج - لو س 4 = 2 \ 3 .
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د - لو 3 ( 2 س ) – لو 3 ( س – 2 ) = 1
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