دروس مادة الرياضيات للفصل الثاني جذع مشترك علوم و تكنولوجيا سنة اولى ثانوي

‫ﻤﻭﺍﻀﻴﻊ ﺍﻹﺭﺴﺎل ﺍﻟﺜﺎﻨﻲ‬ ‫ﻴﺘﻀﻤﻥ ﻫﺫﺍ ﺍﻹﺭﺴﺎل ﺍﻟﻤﻭﺍﻀﻴﻊ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫• ﻤـﻘـﺩﻤـﺔ‬ ‫• ﺍﻟﻌﺒـﺎﺭﺍﺕ ﺍﻟﺠﺒـﺭﻴـﺔ‬ ‫• ﺍﻟﻤﻌﺎﺩﻻﺕ ﻭ ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ‬ ‫• ﺍﻷﺸﻌﺔ – ﺍﻟﻤﻌﺎﻟﻡ – ﺍﻟﻤﺴﺘﻘﻴﻡ‬ ‫• ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ‬ ‫• ﺍﻟـــﺩﻭﺍل‬

‫ﺍﻟﻤـﻘـﺩﻤـﺔ‬‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫• ﻤﺎﺩﺓ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ‬ ‫• ﺍﻟﺒﺭﻨﺎﻤﺞ ﺍﻟﺠﺩﻴﺩ‬ ‫• ﺤل ﺍﻟﻤﺸﻜﻼﺕ‬ ‫• ﺍﻟﺘﺭﻤﻴﺯ ) ﺃﻤﺜﻠﺔ ﺘﻭﻀﻴﺤﻴﺔ (‬

‫• ﻤﺎﺩﺓ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ‪:‬‬‫ﺍﻟﺒﻨﻴﺔ ﺍﻟﺭﻴﺎﻀﻴﺔ ﺘﺄﺜﻴﺭ ﻋﻠﻰ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻔﻜﺭﻴﺔ ﻷﻥ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﻭﺴﻴﻠﺔ ﻟﺘﻜﻭﻴﻥ ﺍﻟﻔﻜﺭ ﻭﺃﺩﺍﺓ ﻻﻜﺘﺴﺎﺏ ﺍﻟﻤﻌﺎﺭﻑ‬ ‫‪.‬ﺘﺴﺎﻫﻡ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﻤﻊ ﺍﻟﻤﻭﺍﺩ ﺍﻷﺨﺭﻯ ﻓﻲ ﺤﺼﺭ ﻤﻠﻤﺢ ﺍﻟﺘﻠﻤﻴﺫ ﻋﻨﺩ ﻨﻬﺎﻴﺔ ﻤﺭﺤﻠﺔ ﺩﺭﺍﺴﻴﺔ ‪.‬‬ ‫• ﺍﻟﺒﺭﻨﺎﻤﺞ ﺍﻟﺠﺩﻴﺩ‬‫ﻜﻤﺎ ﺠﺎﺀ ﺍﻟﺒﺭﻨﺎﻤﺞ ﺍﻟﺭﺴﻤﻲ ‪ ،‬ﺘﻡ ﺒﻨﺎﺀ ﺒﺭﻨﺎﻤﺞ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﻟﻠﺴﻨﺔ ﺍﻷﻭﻟﻰ ﻤﻥ ﺍﻟﺘﻌﻠﻴﻡ ﺍﻟﺜﺎﻨﻭﻱ ) ﺠﺫﻉ ﻤﺸﺘﺭﻙ‬‫ﻋﻠﻭﻡ ﻭ ﺘﻜﻨﻭﻟﻭﺠﻴﺎ ( ﻜﻤﺎ ﻫﻭ ﺍﻟﺸﺄﻥ ﺒﺒﺎﻗﻲ ﺍﻟﻤﻭﺍﺩ ﻭﻓﻕ ﺍﻟﻤﻘﺎﺭﺒﺔ ﺒﺎﻟﻜﻔﺎﺀﺍﺕ ﻭ ﻤﻥ ﺃﺒﻌﺎﺩﻫﺎ ﺍﻷﺴﺎﺴﻴﺔ ﺍﻟﺘﻜﻴﻴﻑ‬ ‫ﻤﻊ ﺍﻟﺠﺩﻴﺩ ﻭﺍﻟﻭﺍﻗﻊ ﺍﻟﻤﻌﺎﺼﺭ ‪ ،‬ﺍﻟﻤﺘﻐﻴﺭ ﺒﺎﺴﺘﻤﺭﺍﺭ ‪.‬‬ ‫• ﺤل ﺍﻟﻤﺸﻜﻼﺕ‪:‬‬ ‫ﺇﻥ ﻤﻜﺎﻨﺔ ﺤل ﺍﻟﻤﺸﻜﻼﺕ ﻓﻲ ﺘﻌﹼﻠﻡ ﺍﻟﺭﻴﺎﻀﻴﺎﺕ ﺃﻤﺭ ﺠﻭﻫﺭﻱ ﻓﻲ ﺒﻨﺎﺀ ﺍﻟﻤﻌﺭﻓﺔ ‪.‬‬ ‫ﻭ ﻤﻥ ﻫﺫﺍ ﺍﻟﻤﻨﻅﻭﺭ ‪ ،‬ﻴﻜﻭﻥ ﺭﺒﻁ ﺍﻟﻤﻌﺎﺭﻑ ﺒﺎﻟﻭﻀﻌﻴﺎﺕ ﻴﺴﻤﺢ ﻟﻠﻤﺘﻌﻠﻡ ﺒﺎﻟﺘﺄﺜﻴﺭ ﺩﺍﺨل ﻭ ﺨﺎﺭﺝ ﺍﻟﻤﺩﺭﺴﺔ ‪.‬‬

‫• ﺍﻟﺘﺭﻤﻴﺯ ‪ ) :‬ﺃﻤﺜﻠﺔ ﺘﻭﻀﻴﺤﻴﺔ (‬ ‫* ﺍﺭﺴﻡ ﻤﺭﺒﻌﺎ ‪ ABCD‬ﺤﻴﺙ ‪. AB = 4 Cm :‬‬ ‫* ﺍﻟﻜﺘﺎﺒﺔ ) ‪ ( x-y‬ﺘﻌﻨﻲ ) ‪ x‬ﻨﺎﻗﺹ ‪. ( y‬‬‫* ﺤل ﻓﻲ ‪ R‬ﺍﻟﻤﻌﺎﺩﻟﺔ ﺫﺍﺕ ﺍﻟﻤﺠﻬﻭل ‪ x‬ﺍﻟﺘﺎﻟﻴﺔ ‪2 x – 1 = 0 :‬‬ ‫* ‪( a + b )2 = a2 + 2 a b + b2‬‬ ‫ﻤﻼﺤﻅﺔ ‪:‬‬ ‫ﺍﻟﻘﺭﺍﺀﺓ ﺍﻟﻨﺴﺒﻴﺔ ‪.‬‬ ‫ﺍ ِﻻﺘﺠﺎﻩ ﺍﻟﻌﺎﻡ ‪.‬‬

‫ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺠـﺒـﺭﻴـﺔ‬ ‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫• ﺘﻌﺭﻴﻑ‬ ‫• ﻨﺸﺭ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ‬ ‫• ﺘﺤﻠﻴل ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ‬ ‫• ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ‬‫• ﺘﺤﻠﻴل ﺍﻟﻌﺒﺎﺭﺓ ‪ax2 + bx + c ; a ≠ o‬‬ ‫• ﺘـﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸـﻜـﻼﺕ‬ ‫• ﺍﻟـﺤـﻠــــﻭل‬

‫• ﺘﻌﺭﻴﻑ ‪:‬‬ ‫‪ - -‬ﻨﺴﻤﻲ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ﺫﺍﺕ ﻤﺘﻐﻴﺭ ﻭﺍﺤﺩ ﻜل ﻋﺒﺎﺭﺓ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻤﺘﻐﻴﺭ ﻭﺍﺤﺩ‪x‬‬‫ﻤﺜﺎل ‪ p(x) = −4x2 + 5x + 3 :‬ﻫﻲ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ﺫﺍﺕ ﻤﺘﻐﻴﺭ ‪. x‬‬ ‫‪ -‬ﻨﺴﻤﻲ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ﺫﺍﺕ ﻤﺘﻐﻴﺭﻴﻥ ﻜل ﻋﺒﺎﺭﺓ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻤﺘﻐﻴﺭﻴﻥ ‪y,x‬‬‫ﻤﺜﺎل ‪ 4x2y − x + 4xy − 5 :‬ﻫﻲ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ﺫﺍﺕ ﻤﺘﻐﻴﺭﻴﻥ ‪y,x‬‬ ‫• ﻨﺸﺭ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ‪:‬‬‫ﻴﻌﺘﻤﺩ ﻨﺸﺭ ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ﻋﻠﻰ ﺍﻟﻘﻭﺍﻋﺩ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪* c(a+b)=ca+cb‬‬‫‪* ( a + b ) ( c + d ) = a c + a d + b c + bd‬‬‫‪* ( a + b )2 = a2 + 2 a b + b2‬‬‫‪* ( a - b )2 = a2 - 2 a b + b2‬‬ ‫ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﺸﻬﻴﺭﺓ‬‫‪* ( a + b ) ( a - b ) = a2 - b2‬‬‫ﻤﺜﺎل ‪ :‬ﺍﻨﺸﺭﺍ ﻟﻌﺒﺎﺭﺓ ﺍﻟﺘﺎﻟﻴﺔ ‪p(x) = 4x(x − 4) − (x + 3)2 :‬‬ ‫ﺍﻟﺤل ‪p(x ) = 4x 2 − 16x − x 2 − 6x − 9 :‬‬ ‫ﻭﻤﻨﻪ ‪. p(x) = 3x2 − 22x − 9 :‬‬ ‫ﻭﻫﻭ ﺍﻟﺸﻜل ﺍﻟﻤﺒﺴﻁ ﻭﺍﻟﻤﺭﺘﺏ ﺤﺴﺏ ﺍﻟﻘﻭﻯ ﺍﻟﻤﺘﻨﺎﻗﺼﺔ ﻟﻠﻌﺩﺩ ‪x‬‬

‫• ﺘﺤﻠﻴل ﻋﺒﺎﺭﺓ ﺠﺒﺭﻴﺔ ‪:‬‬ ‫ﻭﻫﻭ ﻜﺘﺎﺒﺘﻬﺎ ﻋﻠﻰ ﺸﻜل ﺠﺩﺍﺀ ‪ ،‬ﻭ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﺍﻟﻘﻭﺍﻋﺩ ﺍﻵﺘﻴﺔ ‪:‬‬ ‫)‪* ca+cb=c(a+b‬‬ ‫) ‪* a c + a d + b c + bd = ( a + b ) ( c + d‬‬ ‫‪* a2 + 2 a b + b2 = ( a + b )2‬‬ ‫‪* a2 - 2 a b + b2 = ( a - b )2‬‬‫) ‪ * a2 - b2 = ( a + b ) ( a - b‬ﻣﺜﺎل ‪ :‬ﺤﻠل ﺍﻟﻌﺒﺎﺭﺓ‬ ‫ﺍﻟﺘﺎﻟﻴﺔ ‪p(x) = 5(x2 − 4) − (2x − 4)(x + 3) :‬‬ ‫ﺍﻟﺤل ‪ :‬ﻟﺩﻴﻨﺎ ‪p(x) = (x − 2)[5x + 10 − 2x − 6] :‬‬ ‫ﻭﻤﻨﻪ ‪ p(x) = (x − 2)(3x + 4) :‬ﻭﻫﻭ ﺍﻟﺸﻜل ﺍﻟﻤﺤﻠل ﻟﻠﻌﺒﺎﺭﺓ )‪p(x‬‬ ‫• ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ‪:‬‬ ‫ﻜﺘﺎﺒﺔ ﺍﻟﻌﺒﺎﺭﺓ ‪ ax2 + bx + c ; a ≠ o‬ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ‬ ‫ﻭ ﻫﻭ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪b‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪b2 - 4a‬‬ ‫‪c‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2a‬‬ ‫‪‬‬ ‫‪4a2‬‬ ‫‪‬‬ ‫‪a‬‬ ‫‪x2‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫=‬ ‫‪a‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪‬‬ ‫ﻨﻀﻊ ‪ ∆ = b 2 - 4 a c :‬ﻭ ﻴﺴﻤﻰ ﻤﻤﻴﺯ ﺍﻟﻌﺒﺎﺭﺓ ‪. a x2 + b x + c‬‬ ‫ﻤﺜﺎل ‪ :‬ﺍﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ ﺍﻟﻌﺒﺎﺭﺓ ‪p(x) = −4 x2 + 5x + 3 :‬‬‫)‪p(x‬‬ ‫=‬ ‫‪−4‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪5‬‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪(5)2‬‬ ‫)‪− 4(−4)(3‬‬ ‫‪‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪‬‬ ‫)‪× (−4‬‬ ‫‪‬‬ ‫‪4(−4)2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬‫وهﻮ اﻟﺸﻜﻞ اﻟﻨﻤﻮذﺟﻲ ﻟﻠﻌﺒﺎرة )‪p(x‬‬ ‫)‪p(x‬‬ ‫=‬ ‫‪−‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪5 2‬‬ ‫‪−‬‬ ‫‪73 ‬‬ ‫‪‬‬ ‫‪8 ‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪64‬‬ ‫‪‬‬

‫• ﺘﺤﻠﻴل ﺍﻟﻌﺒﺎﺭﺓ ‪ax2 + bx + c ; a ≠ o :‬‬ ‫‪a x2‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫=‬ ‫‪a‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪ 2‬‬ ‫‪-‬‬ ‫∆‬ ‫‪2‬‬ ‫‪‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪‬‬ ‫‪2a‬‬ ‫‪‬‬ ‫‪4a‬‬ ‫‪‬‬ ‫‪ ‬‬ ‫‪‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ < 0 :‬ﻓﺈﻥ ﺍﻟﻌﺒﺎﺭﺓ ‪ a x2 + b x + c‬ﻻ ﺘﺤﻠل‬ ‫‪a ‬‬ ‫‪b‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2a‬‬ ‫‪‬‬‫‪a x2‬‬ ‫‪+bx+c‬‬ ‫=‬ ‫‪x+‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ = 0 :‬ﻓﺈﻥ ‪:‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ > 0 :‬ﻓﺈﻥ ‪:‬‬‫‪a‬‬ ‫‪x2‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫=‬ ‫‪a‬‬ ‫‪ x‬‬ ‫‪−‬‬ ‫‪-‬‬ ‫‪b-‬‬ ‫∆‬ ‫‪‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪-‬‬ ‫‪b‬‬ ‫‪+‬‬ ‫‪∆ ‬‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫‪x2‬‬ ‫=‬ ‫‪−b+‬‬ ‫∆‬ ‫َو‬ ‫‪x1‬‬ ‫=‬ ‫‪−‬‬ ‫‪b−‬‬ ‫∆‬ ‫ﺒﻭﻀﻊ ‪:‬‬ ‫‪2a‬‬ ‫‪2a‬‬‫) ‪a x 2 + b x + c = a ( x − x1 )( x − x 2‬‬ ‫ﻨﺠﺩ ‪:‬‬ ‫ﻤﺜﺎل ‪ :‬ﺤﻠل ﻜل ﻤﻥ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺇﻥ ﺃﻤﻜﻥ ‪.‬‬‫‪p(x) = 5x2 −14x − 3 (2‬‬ ‫‪p(x) = 3x 2 + 2x + 1 (1‬‬ ‫‪p(x) = 4x2 + 4x + 1 (3‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪ p(x) = 3x2 + 2x + 1 (1‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ‪∆ = b2 − 4ac :‬‬‫ﻭﻤﻨﻪ ‪ ∆ = (2)2 − 4(3) (1) :‬ﻭﻋﻠﻴﻪ ‪ ∆ = −8 :‬ﻭﻋﻠﻴﻪ ‪∆ < 0 :‬‬ ‫ﻭﻤﻨﻪ ﻻ ﻴﻤﻜﻥ ﺘﺤﻠﻴل )‪. p(x‬‬ ‫‪ p(x) = 5x2 −14x − 3 (2‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ‪∆ = b2 − 4ab :‬‬ ‫ﻭﻤﻨﻪ ‪ ∆ = (−14)2 − 4(5) (-3) :‬أي ‪∆ = 256 :‬‬

x2 = −b + ∆ ، x1 = −b − ∆ : ‫ﻭﻋﻠﻴﻪ‬ 2a 2a x2 = 14 + 16 =3 ، x1 = 14 - 16 = -1 : ‫ﺃﻱ‬ 2(5) 2 (5) 5 . p(x) = 5  x + 1  ( x-3) : ‫ﻭﻋﻠﻴﻪ‬  5  ∆ = b2 − 4ac :‫ ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ‬p(x) = 4x2 − 4x + 1 (3 ∆ = 0 : ‫( = ∆ ﺃﻱ‬−4)2 − 4 (4) (1) : ‫ﻭﻤﻨﻪ‬. p(x ) = 5  x − 7 2 : ‫ﺃﻱ‬ p(x) = 5 x − 14 2 : ‫ﻭﻋﻠﻴﻪ‬  5  10 

:‫• ﺘـﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸـﻜـﻼﺕ‬ : 1 ‫ﺘﻤﺭﻴﻥ‬ : ‫ﺒﺴﻁ ﻭﺍﻨﺸﺭ ﻤﺎ ﻴﻠﻲ‬A =  5 ax 2  ×  abx 4 ×  b2 x2 3 (1  2   3       5  (2B=( 2 x - 6) ×  1 x5 - 3 x3 +4 x 2 − 1 x 4 + 3 x5 − 3x 4 + 5   2 2 2  (3 ( )( )C = x2 − 4x + 8 - 4x3 + 5x2 − 2x3 + 5x − 3 + 4x2 : 2 ‫ﺘﻤﺭﻴﻥ‬ : ‫ ﺤﻴﺙ‬p(x) , Q(x) ‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺒﺎﺭﺘﻴﻥ‬ Q (x) = - x 2 +5 x-1 , p(x) = 3x 4 − 1 x2 + 5 2p(x) × Q(x) , 2 p (x) - 5 Q(x) , p(x) + Q(x) : ‫ﺍﺤﺴﺏ ﻤﺎ ﻴﻠﻲ‬ : 3 ‫ﺘﻤﺭﻴﻥ‬1 ) a ( x + y )2 = ( ax + ay )2 : ‫ﺃﺫﻜﺭ ﺼﺤﺔ ﺃﻭ ﺨﻁﺄ ﻤﺎ ﻴﻠﻲ‬ .( )2 ) a ( x - y )2 = a 2x - a 2 y 2 .3 ) ( 2x - 3 )2 = 4x 2 + 9 .4) x2 + 4=( x + 2 )( x + 2 ) .5 ) ( 2a - b )2 = 4a 2 - 4ab + b2 .6 ) ( a + b )2 = ( a + b )( a + b ) .

‫ﺘﻤﺭﻴﻥ ‪: 4‬‬‫(‬ ‫‪a - b)3‬‬ ‫=‬ ‫)‪(a - b‬‬ ‫(‬ ‫‪)a -b‬‬ ‫‪2‬‬ ‫‪ (1‬ﻨﻌﻠﻡ‬ ‫ﺃﻥ ‪:‬‬ ‫ﺍﻨﺸﺭ ﺍﻟﻌﺒﺎﺭﺓ ‪ a − b 3‬ﺜﻡ ﺒﺴﻁﻬﺎ ﻭ ﺍﺨﺘﺼﺭﻫﺎ ‪( ).‬‬‫‪ (2‬ﺍﻨﺸﺭ ﺍﻟﻌﺒﺎﺭﺘﻴﻥ ‪( ) ( )(a + b) a2 − ab + b2 , (a − b) a2 +ab +b2‬‬ ‫ﻤﺎﺫﺍ ﺘﺴﺘﻨﺘﺞ ؟‬‫‪ (3‬ﺍﻨﺸﺭ ﻭ ﺍﺨﺘﺼﺭ ﺍﻟﻌﺒﺎﺭﺓ ‪ a + b + c 2‬ﺛﻢ اﺳﺘﻨﺘﺞ ‪( ) ( )4x2 − x + 1 2 :‬‬ ‫ﺘﻤﺭﻴﻥ ‪ : 5‬ﺤﻠل ﺇﻟﻰ ﺠﺩﺍﺀ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪* f (x) = (x - 3)2 -( 2 x - 6) (5 x + 4) -( )9 - x2‬‬‫‪* g (x) =16 (5 x - 3)2 - 25 ( 3 x - 1)2‬‬‫‪* h (x) = 16 x2 + 2 x +‬‬ ‫‪1‬‬ ‫‪16‬‬‫‪* ϕ ( x) = 18 x2 ( x - 4)2 - 8 x4‬‬‫)‪* M ( x , y) = y2 -( x - 1)2 - 2 ( x - 1)( x - y -1‬‬‫‪* Q ( x , y) = a2x y + ab y2 + b2 x y + ab x2‬‬‫‪* T ( x , y ) = x2 + y2 +2x y - 9 a2 + 6ab - b2‬‬ ‫‪f ( x ) = 6 x2 - x -1‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 6‬‬‫؛ ‪Q ( x ) = 3 x2 + x +12‬‬ ‫ﺃﻜﺘﺏ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻋﻠﻰ ﺸﻜﻠﻬﺎ ﺍﻟﻨﻤﻭﺫﺠﻲ ‪:‬‬ ‫‪p ( x ) = x2 + 2 x + 1‬‬ ‫‪39‬‬ ‫* ﺤﻠل ﺇﻟﻰ ﺠﺩﺍﺀ ﻜل ﻤﻥ ﻫﺫﻩ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺇﻥ ﺃﻤﻜﻥ ‪.‬‬‫* ﺍﺴﺘﻨﺘﺞ ﺤﻠﻭل ﻜل ﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ‪َ f ( x ) = 0 :‬ﻭ ‪َ p(x) = 0‬ﻭ ‪Q(x) = 0‬‬

:‫• ﺍﻟـﺤـﻠــــﻭل‬  5 a x2   a4 b4 x4  b6 x6  : 1 ‫ﺘﻤﺭﻴﻥ‬A=  2   81  125      (1 A= 5 a × a4 b4 × b6 × x2 × x4 × x6 2 × 81 × 125 5 a5 b10 x12 . A = 20250 B = x6 - 6x4 + 8x3 - x5 + 3x6 - 6x5 + 10x – 3x5 (2 + 18x3 – 24x2 + 3x4 – 9x5 + 18x4 – 30 . B = 4x 6 – 19x 5 + 15x 4 + 26x 3 – 24x 2 + 10x – 30 C = - 4x 5 + 5x 4 – 2x 5 + 5x 3 – 3x 2 + 4x 4 + 16x 4 – 20x 3 ( 3 + 8x 4 – 20x 2 + 12x – 16x 3 – 32x 3 + 40x2 – 16x3 + 40x – 24 + 32x 2 . C = - 6x 5 + 33x 4 – 79x 3 + 49x 2 + 52x – 24 : 2 ‫ﺘﻤﺭﻴﻥ‬ : p ( x ) + Q ( x ) ‫ ( ﺤﺴﺎﺏ‬1 p ( x ) + Q ( x ) = 3x3 - 1 x2 + 5 - x2 + 5x - 1 2 . p ( x) + Q (x ) = 3x4 - 3 x2 + 5x + 4 2 : 2 P ( x ) – 5 Q ( x ) ‫ ( ﺤﺴﺎﺏ‬2

( )2 P ( x ) - 5 Q ( x ) = 2 3x 4 1 x2   - 2 +5  -5 - x3 + 5x - 1 = 6x 4 – x 2 + 10 + 5x 3 – 25x + 5 . 2 P ( x ) – 5 Q ( x ) = 6x 4 + 5x 3 – x 2 – 25x + 15 : P ( x ) × Q ( x ) ‫ ( ﺤﺴﺎﺏ‬3 ( )P(x)  4 1  × Q(x)=  3x - 2 x 2 +5  -x2 + 5x - 1 = −3x6 + 15x5 − 3x 4 + 1 x4 − 5 x3 + 1 x2 − 5x2 + 25x − 5 2 2 2. p(x) × Q(x) = −3x6 + 15x5 − 5 x4 − 5 x3 − 9 x2 + 25x − 5 2 2 2 : 3 ‫ﺘﻤﺭﻴﻥ‬ . 8 (4 . 8 (3 . 9 (2 . 8 (1 . 9 (7 . 9 (6 . 9 (5 : 4 ‫ﺘﻤﺭﻴﻥ‬ ( a – b ) 3 : ‫ ( ﻨﺸﺭ ﻭﺘﺒﺴﻴﻁ ﻭﺍﺨﺘﺼﺎﺭ ﺍﻟﻌﺒﺎﺭﺓ‬1( )* ( a - b )3 = ( a - b ) ( a - b )2 = ( a - b ) a2 - 2ab + b2 = a3 - 2a2b + ab2 - a2b + 2ab2 - b3 . ( a - b )3 = a3 - 3a2b + 3ab2 - b3 : ‫ ( ﺍﻟﻨﺸﺭ‬2( )* ( a - b ) a2 + ab + b2 = a3 + a2b + ab2 - a2b - ab2 - b3 ( ). ( a - b ) a2 + ab + b2 = a3 - b3

( )* ( a + b ) a2 - ab + b2 = a3 - a2b + ab2 + a2b - ab2 + b3( ). ( a + b ) a2 - ab + b2 = a3 + b3 : ‫ﺍﻻﺴﺘﻨﺘﺎﺝ‬( )* a3 - b3 = ( a - b ) a2 + ab + b2( )* a3 + b3 = ( a + b ) a2 - ab + b2 : ‫ ( ﺍﻟﻨﺸﺭ ﻭﺍﻻﺨﺘﺼﺎﺭ‬3( a + b + c )2 = [ a + ( b + c ) ]2 = a2 + 2a ( b + c ) + ( b + c )2= a2 + 2ab + 2ac + b2 + c2 + 2bc= a2 + b2 + c2 + 2ab + 2ac + 2bc( ): 24x2 - x + 1 ‫ﺍﺴﺘﻨﺘﺎﺝ‬( )4x2 - x + 1 2 =  4x2 + ( - x ) + 1 2 ( ) ( ) ( )= 4x2 2 +(-x)2 +(1)2 +2 4x2 (-x) +2 4x2 (1) +2(-x)(1) = 16x4 + x2 + 1 - 8x3 + 8x2 - 2x( ). 4x2 - x + 1 2 = 16x4 - 8x3 + 9x2 - 2x + 1 : 5 ‫ﺘﻤﺭﻴﻥ‬ ‫ﺍﻟﺘﺤﻠﻴل‬* f ( x ) = ( x - 3 )2 - 2( x - 3 ) ( 5x + 4 ) - ( 3 - x ) ( 3 + x ) = ( x – 3 ) 2 – 2( x – 3 ) ( 5x + 4 ) + ( x – 3 ) ( x + 3 ) = ( x - 3 ) [ x - 3 - 2( 5x + 4 ) - ( x + 3 ) ] = ( x – 3 ) ( x – 3 – 10x – 8 – x – 3 ) = ( x – 3 ) ( - 10x – 14 ) . f ( x ) = - 2( x – 3 ) ( 5x + 7 )

* g ( x ) = [ 4 ( 5x - 3 ) ]2 - [ 5 ( 3x - 1 ) ]2 = [ 20x - 12 ]2 - ( 15x - 5 )2= [( 20x - 12 ) - ( 15x - 5 )][( 20x - 12 ) + ( 15x - 5 )] . g ( x ) = ( 5x - 7 ) ( 35x - 17 )* h ( x ) = ( 4x )2 + 2( 4x ) × 1 +  1 2 =  4x + 1 2 4  4   4 * ϕ (x) = 2x2 9 ( x - 4 )2 - 4x2  = 2x2 ( 3x - 12 )2 - ( 2x )2 = 2x2 [ ( 3x - 12 ) - ( 2x ) ] [ ( 3x - 12 ) + ( 2x ) ] . ϕ ( x ) = 2x2 ( x - 12 ) ( 5x - 12 )* M ( x , y ) = [ y - (x -1)][ y + (x -1)] - 2 ( x - 1 ) ( x - y - 1 ) = (- x + y +1)( x + y -1) - 2 ( x - 1 ) ( x + y - 1 ) = - (x - y -1)( x + y -1) - 2 ( x - 1 ) ( x + y - 1 ) =(x-y-1) [ -x-y+1-2(x-1)] . M(x,y)=(x-y-1)(-3x+y+3) * Q ( x , y ) = a2xy + aby2 + b2xy + abx2 = ay ( ax + by ) + bx ( by + ax ) . Q ( x , y ) = ( ax + by ) ( ay + bx ) * T ( x , y ) = x2 + y2 + 2xy - ( 9a2 - 6ab + b2 ) = ( x + y )2 - ( 3a - b )2

= [ ( x + y ) - ( 3a - b ) ] [ ( x + y ) + ( 3a - b ) ]. T ( x , y ) = ( x + y - 3a + b ) ( x + y + 3a - b ) : 6 ‫ﺘﻤﺭﻴﻥ‬ : ‫ﺍﻟﺸﻜل ﺍﻟﻨﻤﻭﺫﺠﻲ‬ (1f(x)=6   x + - 1 2 - ( - 1 )2 - 4 ( 6 ) ( - 1 )    2 × 6  4 ( 6 )2    . f ( x ) = 6   x - 1 2 - 25    12    144  (2   2 2  2 2 -4 ( 1 )  1       3   9  p(x)=1   x + 3  -    2 )  4 ( 1 )2    (1   . P(x)=  x+ 1 2  3   (3Q(x)=3   x + 1 2 - ( 1 )2 - 4 × 3 × 12    2(3)  4 ( 3 )2    = 3   x + 1 2 + 144    6  36    = 3   x + 1 2 + 4    6    

‫‪f (x) = 6‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪5‬‬ ‫‪‬‬ ‫ﺍﻟﺘﺤﻠﻴل ‪* :‬‬ ‫‪‬‬ ‫‪12‬‬ ‫‪12‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪12‬‬ ‫‪12‬‬ ‫‪‬‬ ‫‪f (x) = 6‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫= )‪p ( x‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫*‬ ‫‪‬‬ ‫‪3 ‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫* ) ‪ Q ( x‬ﻻ ﺘﺤﻠل ‪.‬‬ ‫* ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪: f ( x ) = 0 :‬‬ ‫= )‪f ( x‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪0‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫أو‬ ‫‪x−‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺇﻤﺎ ‪:‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫=‪x‬‬ ‫‪-1‬‬ ‫ﺃﻭ‬ ‫=‪x‬‬ ‫‪1‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪3‬‬ ‫‪2‬‬‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪0‬‬ ‫‪ p ( x) = 0‬ﺃﻱ ‪:‬‬ ‫* ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬‫‪‬‬ ‫‪3‬‬ ‫‪ ‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫ﺃﻭ‬ ‫‪x+‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻨﻪ ﺇﻤﺎ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫=‪x‬‬ ‫‪-1‬‬ ‫ﺃﻭ‬ ‫=‪x‬‬ ‫‪-1‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫* ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ Q x = 0 :‬ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ‪( ).‬‬

‫ﺍﻟﻤﻌﺎﺩﻻﺕ ﻭ ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ‬‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫• ﺤـل ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫• ﺤـل ﻤﺘﺭﺍﺠﺤﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ‬ ‫• ﺘــﺭﻴﻴــﺽ ﺍﻟﻤﺸﻜـﻼﺕ‬‫• ﺍﻟﺤـل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻤﻌﺎﺩﻻﺕ ﺃﻭ ﻤﺘﺭﺍﺠﺤﺎﺕ‬ ‫• ﺘـﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸـﻜـﻼﺕ‬ ‫• ﺍﻟـﺤـﻠــــﻭل‬

‫• ﺤـل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬ ‫‪ax2 + bx + c = 0 ; a ≠ o .1‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ < 0 :‬ﻓﺈﻨﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭﻻ ﻓﻲ ‪. R‬‬ ‫‪x0‬‬ ‫=‬ ‫‪−b‬‬ ‫ﻓﺈﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻼ ﻤﻀﺎﻋﻔﺎ ﻫﻭ ‪:‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪∆ = 0 :‬‬ ‫‪2a‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ > 0 :‬ﻓﺈﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ﻫﻤﺎ ‪:‬‬ ‫‪x2‬‬ ‫=‬ ‫‪−‬‬ ‫‪b+‬‬ ‫∆‬ ‫َﻭ‬ ‫‪x1‬‬ ‫=‬ ‫‪−‬‬ ‫‪b−‬‬ ‫∆‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫ﻤﺜﺎل ‪ :‬ﺤل ﻓﻲ ‪ R‬ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪1 ) x2 - 6x + 9 = 0‬‬‫‪2 ) 3 x2 + 4x + 10 = 0‬‬‫‪3 ) x2 + 4x − 5 = 0‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪ x2 − 6x + 9 = 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 − 4 a c :‬‬‫ﻭﻤﻨﻪ ‪ ∆ = (−6) − 4(9)(1) :‬ﺃﻱ ‪ ∆ = 0 :‬ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ‬ ‫= ‪x0‬‬ ‫‪6 =3‬‬ ‫أي ‪:‬‬ ‫‪x0‬‬ ‫=‬ ‫‪-b‬‬ ‫‪2‬‬ ‫‪2a‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪ 3x2 + 4x + 10 = 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 − 4 a c :‬‬ ‫ﻭﻤﻨﻪ ‪ ∆ = (4)2 − 4(3)(10) = −104 :‬ﺃﻱ ‪∆ < 0 :‬‬ ‫ﻭﻤﻨﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ﻓﻲ ‪. R‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪ x2 + 4x − 5 = 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 − 4ac :‬‬‫ﻭﻤﻨﻪ ‪ ∆ = (4)2 − 4(1) (-5) = 36 :‬أي ‪ ∆ > 0 :‬ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ‬‫= ‪x1‬‬ ‫‪-b-‬‬ ‫∆‬ ‫‪,‬‬ ‫= ‪x2‬‬ ‫‪-b+‬‬ ‫∆‬ ‫ﻤﺘﻤﺎﻴﺯﻴﻥ ‪ x1 , x2‬ﺤﻴﺙ ‪:‬‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫= ‪x1‬‬ ‫‪-‬‬ ‫‪4-‬‬ ‫‪6‬‬ ‫=‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪,‬‬ ‫= ‪x2‬‬ ‫‪- 4 +6‬‬ ‫‪=1‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬

‫• ﺤـل ﻤﺘﺭﺍﺠﺤﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ‪:‬‬ ‫ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ﻭ ﻨﺴﺘﻔﻴﺩ ﻤﻥ ﺇﺸﺎﺭﺓ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻭ ﻫﻲ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ ≤ 0 :‬ﻓﺈﻥ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻤﻥ ﺇﺸﺎﺭﺓ ‪. a‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ ∆ > 0 :‬ﻓﺈﻥ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﺒﻴﻥ ﺍﻟﺠﺫﺭﻴﻥ ﻋﻜﺱ ﺇﺸﺎﺭﺓ ‪ a‬ﻭ ﺨﺎﺭﺝ‬ ‫ﺍﻟﺠﺫﺭﻴﻥ ﻤﻥ ﺇﺸﺎﺭﺓ ‪. a‬‬ ‫ﻤﺜﺎل ‪ :‬ﺤل ﻓﻲ ‪ R‬ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪1 ) x2 - 6x + 9 > 0‬‬‫‪2 ) 3 x2 + 4x + 10 ≤ 0‬‬‫‪3 ) x2 + 4x − 5 < 0‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪ x2 - 6x + 9 > 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 − 4 a c :‬‬ ‫ﻭﻤﻨﻪ ‪ ∆ = (−6)2 − 4(9)(1) :‬ﺃﻱ ‪∆ = 0 :‬‬ ‫ﻭﻤﻨﻪ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ ‪R :‬‬ ‫ﻤﻼﺤﻅﺔ ‪:‬‬ ‫ﻓﻲ ﺤﺎﻟﺔ ‪ x2 - 6x + 9 < 0‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﺨﺎﻟﻴﺔ ‪.‬‬ ‫ﻓﻲ ﺤﺎﻟﺔ ‪ x2 - 6x + 9 ≤ 0‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ ‪{ }. 3 :‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪ 3 x2 + 4x + 10 ≤ 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 − 4 a c :‬‬ ‫ﻭﻤﻨﻪ ‪ ∆ = (4)2 − 4(3)(10) = −104 :‬ﺃﻱ ‪∆ < 0 :‬‬ ‫ﻭﻤﻨﻪ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻤﻭﺠﺏ ﺘﻤﺎﻤﺎ ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻟﻴﺱ ﻟﻬﺎ ﺤل ‪.‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪ x2 + 4x − 5 < 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 − 4ac :‬‬‫ﻭﻤﻨﻪ ‪ ∆ = (4)2 − 4(1) (-5) = 36 :‬ﺃﻱ ‪ ∆ > 0 :‬ﻭﻤﻨﻪ ﻟﺩﻴﻨﺎ ﺠﺫﺭﻴﻥ‬‫= ‪x1‬‬ ‫‪-b-‬‬ ‫∆‬ ‫‪,‬‬ ‫= ‪x2‬‬ ‫‪-b+‬‬ ‫∆‬ ‫ﻤﺘﻤﺎﻴﺯﻴﻥ ‪ x1 , x2‬ﺤﻴﺙ ‪:‬‬ ‫‪2a‬‬ ‫‪2a‬‬ ‫= ‪x1‬‬ ‫‪-‬‬ ‫‪4-‬‬ ‫‪6‬‬ ‫=‬ ‫‪-5‬‬ ‫‪,‬‬ ‫= ‪x2‬‬ ‫‪- 4 +6‬‬ ‫‪=1‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﺒﻴﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺴﺎﻟﺏ ﻭ ﺨﺎﺭﺝ ﺍﻟﺠﺫﺭﻴﻥ ﻤﻭﺠﺏ‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪] [. -5 , 1 :‬‬

‫• ﺘــﺭﻴﻴــﺽ ﺍﻟﻤﺸﻜـﻼﺕ ‪:‬‬ ‫ﻭﻫﻭ ﺘﻭﻅﻴﻑ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺃﻭ ﺍﻟﻤﺘﺭﺍﺠﺤﺎ ﺕ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺃﻭ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻟﺤل ﻤﺸﻜﻼﺕ ‪.‬‬ ‫ﻤﺜﺎل ‪:‬‬‫ﺤﻘل ﻋﻠﻰ ﺸﻜل ﻤﺜـﻠﺙ ﻁﻭﻟﻲ ﻗﺎﻋﺩﺘﻪ ﻭﺍﺭﺘﻔﺎﻋﻪ ﻋﺩﺩﺍﻥ ﻁﺒﻴﻌﻴﺎﻥ ﻤﺘﺘﺎﺒﻌﺎﻥ ﺤﻴﺙ ﻁﻭل ﺍﻟﻘﺎﻋﺩﺓ ﺃﻜﺒﺭ ﻤﻥ‬ ‫ﻁﻭل ﺍﻻﺭﺘﻔﺎﻉ ﻭﻤﺴﺎﺤﺔ ﻫﺫﺍ ﺍﻟﻤﺜـﻠﺙ ﻫﻲ ‪. 780 m2‬‬ ‫‪ -‬ﺍﺤﺴﺏ ﻁﻭﻟﻲ ﻜل ﻤﻥ ﺍﻟﻘﺎﻋﺩﺓ ﻭﺍﻻﺭﺘﻔﺎﻉ ﻟﻬﺫﺍ ﺍﻟﻤﺜـﻠﺙ‪.‬‬ ‫ﺍﻟﺤل ‪:‬‬‫ﻨﻔﺭﺽ ‪ x‬ﻁﻭل ﺍﻻﺭﺘﻔﺎﻉ ‪ .‬ﻓﻴﻜﻭﻥ ﻁﻭل ﺍﻟﻘﺎﻋﺩﺓ ‪x + 1‬‬‫)‪x (x +1‬‬ ‫=‪S‬‬ ‫)‪x (x +1‬‬ ‫‪:‬‬ ‫ﻭﻋﻠﻴﻪ‬ ‫ﻭﻤﻨﻪ ‪2 = 780 :‬‬ ‫‪2‬‬‫ﻭﻤﻨﻪ‪ x x + 1 = 1560 :‬ﻭﺒﺎﻟﺘﺎﻟﻲ ‪( )x2 + x – 1560 = 0 :‬‬ ‫ﻭﻫﻲ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ‪.‬‬‫‪ ∆ = b2 − 4 a c‬ﻭﻋﻠﻴﻪ ‪∆ = ( 1 )2 - 4(1)(-1560) :‬‬ ‫ﺇﺫﻥ ‪ ∆ = 6241 :‬ﻭﺒﺎﻟﺘﺎﻟﻲ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪:‬‬‫ﻭ‬ ‫=‪x1‬‬ ‫‪-b-‬‬ ‫∆‬ ‫=‬ ‫‪−1 − 6241‬‬ ‫‪2a‬‬ ‫)‪2(1‬‬ ‫‪x2‬‬ ‫=‬ ‫‪−b‬‬ ‫‪+‬‬ ‫∆‬ ‫=‬ ‫‪−1+ 6241‬‬ ‫‪24‬‬ ‫)‪2(1‬‬‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ ) x1 = − 40 :‬ﻤﺭﻓﻭﺽ ( َﻭ ‪x2 = 39‬‬‫ﻭﻤﻨﻪ ‪ x = 39 :‬ﻭﻋﻠﻴﻪ ﻁﻭل ﺍﻻﺭﺘﻔﺎﻉ ﻫﻭ ‪ 39 m‬ﻭﻁﻭل ﺍﻟﻘﺎﻋﺩﺓ ﻫﻭ ‪. 40 m‬‬

‫• ﺍﻟﺤـل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻤﻌﺎﺩﻻﺕ ﺃﻭ ﻤﺘﺭﺍﺠﺤﺎﺕ‪:‬‬‫ﻫﻨﺎﻙ ﻤﻌﺎﺩﻻﺕ ﺃﻭ ﻤﺘﺭﺍﺠﺤﺎﺕ ﺘﺤل ﺠﺒﺭﻴﺎ ﻭﻫﻨﺎﻙ ﻤﻌﺎﺩﻻﺕ ﺃﻭ ﻤﺘﺭﺍﺠﺤﺎﺕ ﻻ ﻴﻤﻜﻥ ﺤﻠﻬﺎ ﺠﺒﺭﻴﺎ ﻭﻋﻠﻴﻪ‬ ‫ﻨﺤﻠﻬﺎ ﺒﻴﺎﻨﻴﺎ ‪.‬‬ ‫‪ -‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ f (x) = k :‬ﻫﻭ ﺇﻴﺠﺎﺩ ﻓﻭﺍﺼل ﻨﻘﻁ ﺘﻘﺎﻁﻊ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ‬ ‫ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪. y = k :‬‬ ‫‪ -‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ f(x) = g(x) :‬ﻫﻭ ﺇﻴﺠﺎﺩ ﻓﻭﺍﺼل ﻨﻘﻁ ﺘﻘﺎﻁﻊ ﺍﻟﺘﻤﺜﻴﻠﻴﻥ ﺍﻟﺒﻴﺎﻨﻴﻴﻥ‬ ‫ﻟﻠﺩﺍﻟﺘﻴﻥ ‪ f‬ﻭ ‪. g‬‬ ‫‪ -‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﻌﺔ ‪ f (x) < k :‬ﻫﻭ ﺩﺭﺍﺴﺔ ﺍﻟﻭﻀﻌﻴﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ )‪ (γ‬ﺍﻟﻤﻤﺜل‬‫ﻟﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻭﺍﻟﻤﺴﺘﻘﻴﻡ ) ∆ ( ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪ y = k‬ﺜﻡ ﺍﺴﺘﻨﺘﺎﺝ ﻗﻴﻡ ‪ x‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ )‪(γ‬‬ ‫ﺘﺤﺕ ) ∆ ( ‪.‬‬ ‫‪ -‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) < g(x) :‬ﻫﻭ ﺩﺭﺍﺴﺔ ﺍﻟﻭﻀﻌﻴﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻟﻠﻤﻨﺤﻨﻴﻴﻥ‬ ‫)‪ (γ‬ﻭ )' ‪ (γ‬ﺍﻟﻤﻤﺜﻠﻴﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ ‪َ f‬ﻭ ‪ g‬ﺜﻡ ﺇﻴﺠﺎﺩ ﻗﻴﻡ ‪ x‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ )‪ (γ‬ﺘﺤﺕ )' ‪. (γ‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﺤل ﻓﻲ ‪ R‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ ) x3 = x + 1 :‬ﺘﻌﻁﻲ ﻗﻴﻡ ﺘﻘﺭﻴﺒﻴﺔ ﻟﻠﺤﻠﻭل (‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺘﻴﻥ ‪ f‬ﻭ ‪ g‬ﺤﻴﺙ ‪َ f (x) = x3 :‬ﻭ ‪g(x) = x + 1‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺤﺎﺴﻭﺏ ﻨﻘﻭﻡ ﺒﺈﻨﺸﺎﺀ ﺍﻟﺒﻴﺎﻨﻴﻥ )‪ (γ‬ﻭ )' ‪ (γ‬ﻟﻠﺩﺍﻟﺘﻴﻥ ‪ f‬ﻭ ‪ g‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‬ ‫)' ‪(γ‬‬ ‫)‪(γ‬‬

‫ﻨﻼﺤﻅ ﻋﻠﻰ ﺍﻟﺸﺎﺸﺔ ﻨﻘﻁﺔ ﺍﻟﺘﻘﺎﻁﻊ ﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻬﺎ ﺒﺎﻟﺴﻬﻡ ﻭ ﻋﻠﻰ ﺠﺎﻨﺏ ﺍﻟﺼﻭﺭﺓ ﺘﻅﻬﺭ ﺇﺤﺩﺍﺜﻴﺎﺕ ﻨﻘﻁﺔ‬ ‫ﺍﻟﺘﻘﺎﻁﻊ ﻭ ﻤﻥ ﺍﻟﺤل ﺒﺎﻟﺘﻘﺭﻴﺏ ﻫﻭ ‪. x = 1,3397 :‬‬ ‫• ﺘـﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸـﻜـﻼﺕ‪:‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 1‬‬ ‫ﺃﺫﻜﺭ ﺼﺤﺔ ﺃﻭ ﺨﻁﺄ ﻤﺎ ﻴﻠﻲ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺭﻤﺯ ‪ 9‬ﻟﻠﺼﺤﺔ ﻭﺍﻟﺭﻤﺯ ‪ 8‬ﻟﻠﺨﻁﺄ ‪:‬‬ ‫‪ 16 (1‬ﻫﻭ ﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ‪. x2 = 4 :‬‬ ‫‪ 10 (2‬ﻫﻭ ﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ‪. - 10 x = 0 :‬‬ ‫‪ 5 (3‬ﻟﻴﺱ ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪. x2 - 10 x = - 25 :‬‬‫‪.‬‬ ‫(‬ ‫‪x‬‬ ‫‪+‬‬ ‫)‪2‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫=‬ ‫‪0‬‬ ‫‪:‬‬ ‫ﻟﻠﻤﻌﺎﺩﻟﺔ‬ ‫ﺤﻠﻴﻥ‬ ‫‪1‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪2 ، 2- (4‬‬‫‪ (5‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x2 - 4 = 0 :‬ﺘﻌﻨﻲ ‪. ( x - 2)( x + 2) = 0 :‬‬ ‫‪ (6‬ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ‪ x( x – 4 ) ، x2 - 4 x = 0 :‬ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﺤﻠﻭل ‪.‬‬ ‫‪ (7‬ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺘﻴﻥ ‪2 x2 - 6 x - 2 = 0 , x2 - 3 x - 1 = 0‬‬‫ﻨﻔﺱ ﻤﺠﻭﻋﺔ ﺍﻟﺤﻠﻭل ‪. .‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 2‬‬ ‫ﺤل ﻓﻲ ‪ R‬ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪1 ) 4 ( - 3 x + 5 )2 - 2 ( 6 x - 10 ) ( 5 x - 1 ) = 0‬‬‫‪2 ) 25 ( x + 4 )2 = 9 ( - 3 x + 5 )2‬‬‫) ‪3 ) x2 - 16 = ( x2 - 8 x + 16 ) - ( 3 x -12 ) ( x + 3‬‬‫‪4 ) 4 ( 4 - x2) - ( x - 2 ) 2 = 0‬‬‫) ‪5 ) 12 x3 - 16 x2 = ( 3 x - 4‬‬‫‪6 ) 2 x2 - 18 - ( x + 3 )2 + 8 x + 24 = 0‬‬

‫ﺘﻤﺭﻴﻥ ‪: 3‬‬ ‫ﺤل ﻓﻲ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺤﻴﺙ ‪ x‬ﻤﺠﻬﻭل َﻭ ‪ m‬ﻭﺴﻴﻁ ﺤﻘﻴﻘﻲ ‪:‬‬‫‪1 ) - 6 x2 + 5 x + 25 = 0‬‬‫‪2 ) 25 x2 + 10 x + 1 = 0‬‬‫‪3 ) - 5 x2 + 3 x - 4 = 0‬‬‫‪4 ) m x2 + 2 (m + 1 ) x + m - 4 = 0‬‬‫‪5 ) m x3 +2 m x2 + m x - 2 x2 - 2 x = 0‬‬‫‪6 ) x2 + 2 ( m - 4 ) x + m2 + 4 = 0‬‬‫) ‪1 ) 8x 2 – 50 = ( 2x – 5 ) ( m x – 1‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 4‬‬‫)‪2) (x–3)2=m(x2–9‬‬‫‪3 ) ( x – 5 ) ( x 2 – 6x + 9 ) ( + x 2 + m x – m 2 ) = 0‬‬ ‫ﺤل ﻓﻲ ‪ R‬ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 5‬‬ ‫ﻗﻁﻌﺔ ﺃﺜﺭﻴﺔ ﻋﻠﻰ ﺸﻜل ﺸﺒﻪ ﻤﻨﺤﺭﻑ ﻁﻭل ﻗﺎﻋﺩﺘﻪ ﺍﻟﻜﺒﺭﻯ ﻴﺴﺎﻭﻱ ﻁﻭل ﺍﺭﺘﻔﺎﻋﻪ‬ ‫ﻭﻁﻭل ﻗﺎﻋﺩﺘﻪ ﺍﻟﺼﻐﺭﻯ ﻨﺼﻑ ﻁﻭل ﺍﺭﺘﻔﺎﻋﻪ ﻭ ﻤﺴﺎﺤﺘﻪ ﻫﻲ ‪48 Cm2‬‬ ‫‪E‬‬ ‫‪ -‬ﺍﺤﺴﺏ ﺃﻁﻭﺍل ﻜل ﻤﻥ ﻗﺎﻋﺩﺘﻴﻪ ﻭ ﺍﺭﺘﻔﺎﻋﻪ ‪.‬‬‫‪x‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 6‬‬ ‫‪A‬‬ ‫ﺍﺤﺴﺏ ‪ x‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻤﺴﺎﺤﺔ ﺍﻟﻤﻨﺯل‬ ‫‪ A B C D E‬ﺘﺴﺎﻭﻱ ‪D 180 m2‬‬‫‪x‬‬ ‫‪B‬‬ ‫‪2x-1‬‬ ‫‪C‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 7‬‬ ‫ﺍﺸﺘﺭﻯ ﻤﺤﻤﺩ ‪ 10‬ﻜﺭﺍﺭﻴﺱ ﻭ ‪ 5‬ﻜﺘﺏ ﺒـ ‪. 2200 DA‬‬‫ﺍﺤﺴﺏ ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ﺍﻟﻭﺍﺤﺩ ﺜﻡ ﺴﻌﺭ ﺍﻟﻜﺘﺎﺏ ﺍﻟﻭﺍﺤﺩ ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ﻤﺭﺒﻊ ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ﺍﻟﻭﺍﺤﺩ ﻴﺴﺎﻭﻱ‬ ‫ﺴﻌﺭ ﺍﻟﻜﺘﺎﺏ ﺍﻟﻭﺍﺤﺩ ‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 8‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﺤﻴﺙ ‪ AB = AC :‬ﻭ‪ H‬ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁﺔ ‪ A‬ﻋﻠﻰ ) ‪.( BC‬‬ ‫ﻁﻭل ‪ BC‬ﻴﺯﻴﺩ ﺒـ ‪ 1 Cm‬ﻋﻥ ﻁﻭل ‪[ ] [ ]. AB‬‬ ‫‪ -‬ﻋﻴﻥ ﻁﻭل ﻜل ﻤﻥ ‪َ AB‬ﻭ ‪ BC‬ﻋﻠﻤﺎ ﺃﻥ ‪[ ] [ ]. AH = 4 Cm :‬‬

‫ﺘﻤﺭﻴﻥ ‪: 9‬‬ ‫ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﺃﻁﻭﺍل ﺃﻀﻼﻋﻪ ﺍﻟﺜﻼﺜﺔ ﻫﻲ ﺃﻋﺩﺍﺩ ﻁﺒﻴﻌﻴﺔ ﻤﺘﺘﺎﺒﻌﺔ ﺍﺤﺴﺏ ﻜل ﻤﻨﻬﺎ ‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 10‬‬ ‫‪15‬‬ ‫ﻋﻨﺩ ﻁﺭﺡ ﻋﺩﺩ ﻁﺒﻴﻌﻲ ﻤﻥ ﻜل ﻤﻥ ﺒﺴﻁ ﻭﻤﻘﺎﻡ ﺍﻟﻜﺴﺭ ‪ 8‬ﻨﺠﺩ ﻋﺩﺩﺍ ﻤﺴﺎﻭﻴﺎ ﺇﻟﻰ‪.8‬‬ ‫‪ -‬ﻤﺎ ﻫﻭ ﻫﺫﺍ ﺍﻟﻌﺩﺩ ؟‬ ‫ﺘﻤﺭﻴﻥ ‪: 11‬‬ ‫ﺃﺭﺍﺩ ﻁﻔل ﺘﺭﺘﻴﺏ ﻤﺠﻤﻭﻋﺔ ﻤﻥ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ﻋﻠﻰ ﺸﻜل ﻤﺭﺒﻊ ﻓﺒﻘﻲ ﻟﻪ ‪ 11‬ﻨﺭﺩ‪ ،‬ﻓﻘﺭﺭ ﺇﻀﺎﻓﺔ ﻨﺭﺩ ﻓﻲ‬ ‫ﻜل ﺼﻑ ﻭ ﻫﺫﺍ ﺃﺩﻯ ﺇﻟﻰ ﻨﻘﺹ ‪ 16‬ﻨﺭﺩ ﻹﺘﻤﺎﻡ ﺍﻟﻤﺭﺒﻊ ﺍﻟﺠﺩﻴﺩ‬ ‫‪.‬ﻤﺎ ﻫﻭ ﻋﺩﺩ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ﺍﻟﺘﻲ ﺍﺴﺘﻌﻤﻠﻬﺎ ﺍﻟﻁﻔـل ‪.‬‬ ‫زهﺮة اﻟﻨﺮد‬ ‫ﺘﻤﺭﻴﻥ ‪: 12‬‬ ‫ﻋﻴﻥ ﺜﻼﺜﺔ ﺃﻋﺩﺍﺩ ﻤﺘﺘﺎﺒﻌﺔ ‪ .‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﻔﺭﻕ ﺒﻴﻥ ﺠﺩﺍﺀ ﺍﻟﻌﺩﺩﻴﻥ ﺍﻷﻜﺒﺭ ﻭ ﻤﺭﺒﻊ ﺍﻟﻌﺩﺩ ﺍﻷﺼﻐﺭ ﻴﺴﺎﻭﻱ‬ ‫‪D‬‬ ‫‪. 41‬‬‫‪A‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 13‬‬ ‫إﻟﻴﻙ ﺍﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل ‪:‬‬ ‫ﻋﻴﻥ ﻭﻀﻊ ﺍﻟﻨﻘﻁﺔ ‪ M‬ﺒﺤﻴﺙ‬ ‫‪8 cm‬‬ ‫‪M 4 cm‬‬ ‫ﺘﻜﻭﻥ ‪. AM = DM :‬‬‫‪B‬‬ ‫‪5 cm‬‬ ‫‪C‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 14‬‬ ‫ﺃﺭﺍﺩ ﺭﻀﺎ ﺸﺭﺍﺀ ‪ 8‬ﺃﻗﻼﻡ ﻭ ‪ 20‬ﻜﺭﺍﺱ ﺤﻴﺙ ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ﻻ ﻴﻘل ﻋﻥ ‪ 20 DA‬ﻭﻫﻭ ﻀﻌﻑ ﺴﻌﺭ‬ ‫ﺍﻟﻘﻠﻡ ﻤﺎ ﻫﻭ ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ﻭ ﺍﻟﻘﻠﻡ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﺒﻠﻎ ﺍﻟﺫﻱ ﺩﻓﻌﻪ ﺭﻀﺎ ﺃﻗـل ﻤﻥ ‪. 1000 DA‬‬

‫ﺘﻤﺭﻴﻥ ‪: 15‬‬ ‫ﺤل ﻓﻲ ‪ R‬ﻜل ﻤﻥ ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫)‪1‬‬ ‫‪−x+3‬‬ ‫≤‬ ‫‪0‬‬ ‫‪2x+5‬‬‫‪2 ) (−5x + 2) (4x + 5) ≥ 0‬‬‫)‪3‬‬ ‫‪2x‬‬ ‫‪-‬‬ ‫‪2x + 4‬‬ ‫≥‬ ‫‪0‬‬ ‫‪x+ 4‬‬ ‫‪x-5‬‬‫‪4 ) 4 x2 − 25 - ( 2 x - 5 ) (x + 3) ≥ 0‬‬‫‪5 ) 20 x2 + 6x - 2 < 0‬‬‫‪6 ) 49 x2 - 14 x + 1 ≤ 0‬‬ ‫ﻭﻤﺤﻴﻁﻪ ‪. 10 Cm2‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 16‬‬ ‫ﻫل ﻴﻭﺠﺩ ﻤﺴﺘﻁﻴل ﻤﺴﺎﺤﺘﻪ ‪6 Cm2‬‬ ‫ﻴﻁﻠﺏ ﺘﻌﻴﻴﻥ ﺍﻟﺤل ﺠﺒﺭﻴﺎ ﺜﻡ ﺒﻴﺎﻨﻴﺎ ‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 17‬‬ ‫ﻟﺘﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ −3 ; 4‬ﺒﺒﻴﺎﻨﻬﺎ ‪ γ‬ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ ‪( ) [ ].‬‬ ‫‪y‬‬ ‫‪ -‬ﺤل ﺒﻴﺎﻨﻴﺎ ﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪y = -x+1‬‬ ‫‪f (x) ≤ 0 (1‬‬ ‫‪f (x) = -1 (2‬‬ ‫) ‪(∆2‬‬ ‫‪f (x) = - x +1 (3‬‬ ‫‪x' -3‬‬ ‫)‪(γ‬‬ ‫‪x‬‬ ‫‪4‬‬ ‫‪y = -1‬‬ ‫)‪y' (∆1‬‬

‫ﺘﻤﺭﻴﻥ ‪: 18‬‬‫‪ ( 1‬ﺤل ﺒﻴﺎﻨﻴﺎ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ 4 - x2 = −5 :‬ﺜﻡ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪4 - x2 ≥ 0 :‬‬ ‫‪ ( 2‬ﺠﺩ ﻫﺫﻩ ﺍﻟﻨﺘﺎﺌﺞ ﺤﺴﺎﺒﻴﺎ ‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 19‬‬‫ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ﺃﻨﺸﺊ ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﺒﻴﺎﻨﻴﺔ ‪( ) ( ) ( )U1 ; U2 ; U3‬‬ ‫ﻟﻠﺩﻭﺍل ‪ h , g , f‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﺤﻴﺙ ‪:‬‬‫‪f ( x ) = x2 - x‬‬ ‫;‬ ‫‪g‬‬ ‫(‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫‪4‬‬ ‫;‬ ‫‪h(x)=x+3‬‬ ‫‪x‬‬ ‫* ﺤل ﺒﻴﺎﻨﻴﺎ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪f ( x ) = - 2 :‬‬ ‫* ﺤل ﺒﻴﺎﻨﻴﺎ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪f ( x ) > - 2 :‬‬ ‫‪4‬‬ ‫≤‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪:‬‬ ‫ﺍﻟﻤﺘﺭﺍﺠﺤﺔ‬ ‫ﺒﻴﺎﻨﻴﺎ‬ ‫ﺜﻡ‬ ‫ﺤﺴﺎﺒﻴﺎ‬ ‫ﺤل‬ ‫*‬ ‫‪x‬‬ ‫* ﺤل ﺤﺴﺎﺒﻴﺎ ﺜﻡ ﺒﻴﺎﻨﻴﺎ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪x + 3 ≥ x 2 - x :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 20‬‬‫ﻟﺘﻜﻥ‪ :‬ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ − 5 ; 5‬ﺒﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ‪ γ‬ﺍﻟﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻰ ‪( ) [ ]:‬‬ ‫‪y‬‬‫‪(γ) 3‬‬ ‫‪2‬‬‫'‪x‬‬ ‫‪1‬‬ ‫‪3,5 5 x‬‬ ‫‪12‬‬ ‫‪-4 -3‬‬ ‫‪-3‬‬ ‫'‪y‬‬‫ﺃﺫﻜﺭ ﺼﺤﺔ ﺃﻭ ﺨﻁﺄ ﻤﺎ ﻴﻠﻲ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺭﻤﺯ ‪ 9‬ﻟﻠﺼﺤﺔ ﻭﺍﻟﺭﻤﺯ ‪ 8‬ﻟﻠﺨﻁﺄ ‪:‬‬ ‫‪ (1‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f (x) = 1‬ﺤﻠﻴﻥ ‪.‬‬ ‫‪ (2‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f (x) = 0‬ﺤﻠﻴﻥ ‪.‬‬

‫‪ (3‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f(x) = - 1‬ﺤﻠﻴﻥ ‪.‬‬ ‫‪ (4‬ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) ≤ 0 :‬هﻲ ‪[ ]. E1 = -5 , -4 :‬‬‫‪ (5‬ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) > 0 :‬هﻲ ‪] [. E2 = - 4 ; 2 :‬‬ ‫‪ (6‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f (x) = x‬ﺤل ﻭﺤﻴﺩ ‪. .‬‬ ‫‪ (7‬ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) < - 3 :‬هﻲ ‪[ ]E3 = -4 ; 2 :‬‬ ‫‪.‬‬ ‫‪ (8‬ﻟﻴﺱ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) f 3‬ﺃﻱ ﺤل ‪. .‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 21‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺤﺎﺴﺒﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ﺃﻨﺸﺊ ﺘﻤﺜﻴﻼ ﺒﻴﺎﻨﻴﺎ ‪ γ‬ﻟﻠﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ‪ f ( x)= Sin x:‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل) (‬ ‫]‪. [−π , π‬‬ ‫‪ -‬ﺍﺴﺘﻨﺘﺞ ﺒﻴﺎﻨﻴﺎ ﻋﻠﻰ ‪ −π , π‬ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ] [‬‫‪1 ) f (x) = - 1‬‬ ‫‪; 2 ) f (x) = 1 ; 3 ) f (x) = 0‬‬ ‫‪ -‬ﺍﺴﺘﻨﺘﺞ ﺒﻴﺎﻨﻴﺎ ﻋﻠﻰ ‪ −π , π‬ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ‪[ ]:‬‬‫‪1 ) f (x) ≤ - 1 ; 2 ) f (x) > 1 ; 3 ) f (x) ≥ 0‬‬ ‫ﻋﻴﻨﻬﺎ ﺒﺘﻘﺭﻴﺏ ‪. 0,001‬‬ ‫‪f‬‬ ‫= )‪(x‬‬ ‫‪1‬‬ ‫ﻤﺎ ﻫﻭ ﻋﺩﺩ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬ ‫‪2‬‬

‫• ﺍﻟـﺤـﻠــــﻭل‪:‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 1‬‬ ‫‪. × (3‬‬ ‫‪. × (2‬‬ ‫‪. × (1‬‬ ‫‪. 9 (4‬‬‫‪. × (7‬‬ ‫‪. 9 (6. 9 (5‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 2‬‬‫‪ ( 1‬ﻟﺩﻴﻨﺎ ‪4 ( - 3 x + 5 )2 - 2 ( 6 x - 10 ) ( 5 x - 1 ) = 0 :‬‬ ‫‪4 ( 3 x - 5 )2 - 4 ( 3 x -5 ) ( 5 x - 1) = 0‬‬ ‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫‪4 (3 x - 5 ) 3 x − 5 − (5 x −1) = 0‬‬ ‫‪4(3x–5)( -2x–4)=0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪-2x–4=0‬‬ ‫‪ 3 x – 5 = 0‬أو‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪x=-2‬‬ ‫أو‬ ‫=‪x‬‬ ‫‪5‬‬ ‫‪3‬‬ ‫‪ ( 2‬ﻟﺩﻴﻨﺎ ‪25 ( x + 4 )2 = 9 ( - 3 x + 5 )2 :‬‬‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪5 ( x + 4)2 − 3 (- 3 x + 5)2 = 0 :‬‬ ‫ﺃﻱ ‪:‬‬‫‪5 (x + 4) − 3(- 3 x + 5) 5(x + 4) + 3(- 3 x + 5) = 0‬‬ ‫ﻭﻤﻨﻪ ‪( 5 x + 20 + 9 x – 15 ) ( 5 x + 20 – 9 x + 15 ) = 0 :‬‬ ‫‪( 14 x + 5 ) ( - 4 x + 35 ) = 0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫ﺇﺫﻥ ‪ 14 x + 5 = 0 :‬أو ‪- 4 x + 35 = 0‬‬ ‫‪x‬‬ ‫=‬ ‫‪35‬‬ ‫أو‬ ‫‪x‬‬ ‫=‬ ‫‪-5‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬ ‫‪4‬‬ ‫‪14‬‬ ‫‪ ( 3‬ﻟﺩﻴﻨﺎ ‪x2 - 16 = ( x2 - 8 x + 16 ) - ( 3 x -12 ) ( x + 3 ) :‬‬‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪x2 - 16 - ( x - 4 )2 + 3( x - 4)( x + 3 ) = 0 :‬‬

‫‪(x-4) [ x+4-(x-4)+3(x+3) ] =0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪( x - 4 ) ( 3x + 17 ) = 0‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫ﺇﺫﻥ ‪ x – 4 = 0 :‬أو ‪3x + 17 = 0‬‬ ‫‪.‬‬ ‫‪x = - 17‬‬ ‫أو‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪x = 4 :‬‬ ‫‪3‬‬ ‫‪ ( 4‬ﻟﺩﻴﻨﺎ ‪4 ( 4 - x2 ) - ( x - 2 ) 2 = 0 :‬‬ ‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪4 ( 2 - x ) ( 2 + x ) - ( x - 2 )2 = 0 :‬‬‫‪(x-2)[-4(x+2)-(x-2)] =0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪( x - 2 ) ( - 5x - 6 ) = 0‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪:‬‬‫‪x= -6‬‬ ‫ﺃﻭ‬ ‫ﺃﻱ ‪x = 2 :‬‬ ‫‪ x - 2 = 0‬ﺃﻭ ‪- 5x - 6 = 0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪5‬‬ ‫‪ ( 5‬ﻟﺩﻴﻨﺎ ‪12 x3 - 16 x2 = ( 3 x - 4 ) :‬‬ ‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪4x2 ( 3x - 4 ) - ( 3x - 4 ) = 0 :‬‬ ‫‪( 3x - 4 ) ( 4x2 - 1 ) = 0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪( 3x - 4 ) ( 2x - 1 ) ( 2x + 1 ) = 0‬‬ ‫ﺃﻱ ‪:‬‬‫ﻭﻋﻠﻴﻪ ‪ 3x - 4 = 0 :‬ﺃﻭ ‪ 2x - 1 = 0‬ﺃﻭ ‪2x + 1 = 0‬‬ ‫‪.‬‬ ‫=‪x‬‬ ‫‪-1‬‬ ‫ﺃﻭ‬ ‫=‪x‬‬ ‫‪1‬‬ ‫ﺃﻭ‬ ‫=‪x‬‬ ‫‪4‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪3‬‬ ‫‪ ( 6‬ﻟﺩﻴﻨﺎ ‪2 x2 - 18 - ( x + 3 )2 + 8 x + 24 = 0 :‬‬‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪( )2 x2 - 9 - ( x + 3 )2 + 8 ( x + 3 ) = 0 :‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2 ( x - 3 ) ( x + 3 ) - ( x + 3 )2 + 8 ( x + 3 ) = 0‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪( x + 3 ) [ 2 ( x - 3 ) - ( x + 3 ) + 8 ] = 0 :‬‬‫ﺃﻱ ‪ ( x + 3 ) ( x - 1 ) = 0 :‬ﺇﺫﻥ ‪ x + 3 = 0 :‬ﺃﻭ ‪x - 1 = 0‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ x = - 3 :‬ﺃﻭ ‪. x = 1‬‬

‫ﺘﻤﺭﻴﻥ ‪: 3‬‬ ‫‪- 6x2 + 5x + 25 = 0 ( 1‬‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = ( 5 )2 - 4 ( - 6 ) ( 25 ) = 625 :‬‬‫‪x2‬‬ ‫=‬ ‫‪-b+‬‬ ‫∆‬ ‫ﻭ‬ ‫‪x1‬‬ ‫=‬ ‫‪-b-‬‬ ‫‪ ∆ > 0‬ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ∆‬ ‫‪2a‬‬ ‫‪2a‬‬‫‪x2‬‬ ‫=‬ ‫‪- 5 + 625‬‬ ‫=‬ ‫‪−5‬‬ ‫ﻭ‬ ‫‪x1‬‬ ‫=‬ ‫‪- 5 - 625‬‬ ‫=‬ ‫‪5‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫) ‪2( - 6‬‬ ‫‪3‬‬ ‫) ‪2( - 6‬‬ ‫‪2‬‬ ‫‪25x2 + 10 x + 1 = 0 ( 2‬‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = ( 10 )2 - 4 ( 1 ) ( 25 ) :‬‬ ‫ﻭﻤﻨﻪ ‪∆= 100 - 100 = 0 :‬‬ ‫‪x0‬‬ ‫=‬ ‫‪-b‬‬ ‫=‬ ‫‪- 10‬‬ ‫=‬ ‫‪-1‬‬ ‫ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ ‪:‬‬ ‫‪2a‬‬ ‫‪50‬‬ ‫‪5‬‬ ‫‪- 5x2 + 3x - 4 = 0 ( 3‬‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = ( 3 )2 - 4 ( - 5 ) ( - 4 ) = - 71 :‬‬ ‫‪ ∆ < 0‬ﻭﻋﻠﻴﻪ ‪ :‬ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ﻓﻲ ‪. R‬‬ ‫‪m x2 + 2 ( m + 1 ) x + m - 4 = 0 ( 4‬‬ ‫* ﻤﻥ ﺃﺠل ‪ : m = 0‬ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪ 2x - 4 = 0 :‬ﻭﻋﻠﻴﻪ ‪. x = 2 :‬‬ ‫* ﻤﻥ ﺃﺠل ‪ : m ≠ 0‬ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac‬‬ ‫) ‪∆ = 4 ( m + 1 )2 - 4 ( m ) ( m - 4‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪( )= 4 m2 + 2m + 1 - 4m2 + 16m‬‬ ‫ﻭﻋﻠﻴﻪ ‪∆ = 24m + 4 :‬‬ ‫∞‪-‬‬ ‫‪-1‬‬ ‫∞‪+‬‬ ‫ﺇﺸﺎﺭﺓ ∆ ‪:‬‬ ‫‪6‬‬ ‫‪m‬‬ ‫ـــ ∆‬ ‫‪+‬‬ ‫* ﻟﻤﺎ ‪:‬‬

‫ﻓﺈﻥ ‪ ∆ < 0 :‬ﻭﻋﻠﻴﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ﻓﻲ ‪. R‬‬ ‫‪m‬‬ ‫∈‬ ‫‪‬‬ ‫‪-∞ ,‬‬ ‫‪-1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪‬‬‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪x0‬‬ ‫=‬ ‫‪-b‬‬ ‫ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ ‪:‬‬ ‫ﻓﺈﻥ ‪∆ = 0 :‬‬ ‫=‪m‬‬ ‫‪-1‬‬ ‫‪2a‬‬ ‫* ﻟﻤﺎ ‪6 :‬‬ ‫ﻭ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ ‪:‬‬ ‫‪x0‬‬ ‫=‬ ‫‪-2(m+‬‬ ‫‪1‬‬ ‫)‬ ‫‪2m‬‬ ‫‪-1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪5‬‬ ‫‪6‬‬ ‫=‪. x0‬‬ ‫‪-1‬‬ ‫=‬ ‫‪6‬‬ ‫‪=−‬‬ ‫‪5‬‬ ‫‪-1‬‬ ‫‪66‬‬ ‫ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ‬ ‫‪∆>0‬‬ ‫ﻓﺈﻥ‬ ‫∈‪m‬‬ ‫‪‬‬ ‫‪-1‬‬ ‫‪,‬‬ ‫‪+‬‬ ‫∞‬ ‫‪‬‬ ‫ﻟﻤﺎ‬ ‫*‬ ‫‪‬‬ ‫‪6‬‬ ‫‪‬‬ ‫‪x1‬‬ ‫=‬ ‫‪-b-‬‬ ‫∆‬ ‫=‬ ‫‪-2(m+1) - 24m+4‬‬ ‫‪2a‬‬ ‫‪2m‬‬ ‫‪x2‬‬ ‫=‬ ‫‪-b+‬‬ ‫=∆‬ ‫‪-2(m+1) + 24m+4‬‬ ‫ﻭ‬ ‫‪2a‬‬ ‫‪2m‬‬ ‫‪m x3 + 2 m x2 + m x - 2 x2 - 2 x = 0 ( 5‬‬‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪(1) . . . x m x2 + 2 m x + m - 2 x - 2  = 0 :‬‬ ‫ﺃﻱ ‪ x = 0 :‬ﺃﻭ ‪x2 + ( 2m - 2 ) x + m - 2 = 0 :‬‬‫‪(2) . . . m x2 + ( 2m - 2 ) x + m - 2 = 0‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬‫‪ ∆ = b2 - 4ac‬ﻭﻋﻠﻴﻪ ‪∆ = ( 2m - 2 )2 - 4 × m ( m - 2 ) :‬‬‫‪∆=4‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﺇﺫﻥ ‪∆ = 4m2 - 8m + 4 - 4m2 + 8m :‬‬ ‫ﺇﺫﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ )‪ (2‬ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ﻷﻥ ‪∆ > 0 :‬‬ ‫‪x1‬‬ ‫=‬ ‫‪−b‬‬ ‫‪−‬‬ ‫∆‬ ‫=‬ ‫‪−(2m − 2) − 2‬‬ ‫=‬ ‫‪-m‬‬ ‫‪2a‬‬ ‫)‪2(1‬‬ ‫‪x2‬‬ ‫=‬ ‫‪−b +‬‬ ‫∆‬ ‫=‬ ‫‪−(2m − 2) + 2 = - m + 2‬‬ ‫ﻭ‬ ‫‪2a‬‬ ‫)‪2(1‬‬

‫‪x2 + 2 ( m – 4 ) x + m2 + 4 = 0 (6‬‬‫‪ ∆ = b2 − 4ac‬ﻭ ﻋﻠﻴﻪ ‪∆ = 4(m − 4)2 − 4(1)(m2 + 4) :‬‬ ‫ﺃﻱ ‪ ∆ = 4m2 − 32 + 64 − 4m2 − 16 :‬وﻋﻠﻴﻪ ‪:‬‬ ‫‪∆ = −32m + 48‬‬ ‫‪m‬‬ ‫∞‪-‬‬ ‫‪3‬‬ ‫ﺇﺸﺎﺭﺓ ∆ ‪:‬‬ ‫‪2‬‬ ‫∞‪+‬‬ ‫‪∆+‬‬ ‫ـــ‬ ‫* ﻟﻤﺎ‬‫‪ ∆ > 0‬ﻭ ﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ‬ ‫‪:‬‬ ‫ﻓﺈﻥ‬ ‫‪m‬‬ ‫∈‬ ‫‪‬‬ ‫∞‪−‬‬ ‫‪,‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪2 ‬‬‫= ‪x1‬‬ ‫‪-b-‬‬ ‫∆‬ ‫=‬ ‫‪-2(m - 4) -‬‬ ‫‪- 32 m + 48‬‬ ‫‪2a‬‬ ‫‪2‬‬‫‪x2‬‬ ‫=‬ ‫‪-b+‬‬ ‫=∆‬ ‫‪-2(m - 4) +‬‬ ‫‪- 32 m + 48‬‬ ‫ﻭ‬ ‫‪2a‬‬ ‫‪2‬‬ ‫‪ ∆ = 0‬ﻭ ﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ ﻫﻭ ‪:‬‬ ‫‪:‬‬ ‫ﻓﺈﻥ‬ ‫=‪m‬‬ ‫‪3‬‬ ‫ﻟﻤﺎ‬ ‫*‬ ‫‪2‬‬ ‫‪-b‬‬ ‫) ‪−2( m - 4‬‬ ‫ﻭ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ ‪:‬‬ ‫‪x0‬‬ ‫=‬ ‫‪2a‬‬ ‫=‬ ‫‪2×1‬‬ ‫‪=4−m‬‬ ‫‪x0‬‬ ‫=‬ ‫‪4‬‬ ‫‪−‬‬ ‫‪3‬‬ ‫=‬ ‫‪5‬‬ ‫‪2‬‬ ‫‪2‬‬‫∈ ‪ m‬ﻓﺈﻥ ‪ ∆ < 0 :‬ﻭﻋﻠﻴﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ﻓﻲ ‪.R‬‬ ‫‪‬‬ ‫‪3‬‬ ‫∞‪,+‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫* ﻟﻤﺎ ‪ :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 4‬‬ ‫‪ ( 1‬ﻟﺩﻴﻨﺎ ‪8x 2 – 50 = ( 2x – 5 ) ( m x – 1 ) :‬‬‫ﺃﻱ ‪2 ( 2 x - 5 ) ( 2 x + 5 ) - (2 x - 5 ) ( m x - 1 ) = 0 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪( 2 x - 5 ) [ ( 4 - m ) x + 11] = 0 :‬‬‫ﻭ ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ‪ 2 x - 5 = 0 :‬ﺃﻭ ‪(4 - m ) x + 11 = 0‬‬ ‫ﺃﻭ ‪( 4 - m ) x + 11 = 0‬‬ ‫=‪x‬‬ ‫‪5‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪2‬‬

‫‪(4 - m ) x + 11 = 0‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m = 4‬ﻟﺩﻴﻨﺎ ‪ O × x + 11 = 0 :‬ﻭﻫﺫﺍ ﻤﺴﺘﺤﻴل ‪.‬‬ ‫=‪x‬‬ ‫‪- 11‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m ≠ 4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪4-m‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫=‪x‬‬ ‫‪5‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m = 4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪5‬‬ ‫‪- 11‬‬ ‫=‪x‬‬ ‫‪2‬‬ ‫ﺃﻭ‬ ‫=‪x‬‬ ‫‪4-m‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m ≠ 4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ ( 2‬ﻟﺩﻴﻨﺎ ‪( x – 3 ) 2 = m ( x 2 – 9 ) :‬‬ ‫ﺃﻱ ‪( x - 3 ) [ x - 3 - m ( x + 3 ) ] = 0 :‬‬ ‫ﺃﻱ ‪( x - 3 ) [ ( 1 - m ) x - 3 - 3 m]= 0 :‬‬ ‫ﻭﻤﻨﻪ ‪ x - 3 = 0 :‬ﺃﻱ ‪ x = 3 :‬ﺃﻭ ‪:‬‬ ‫‪(1-m)x-3-3m=0‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪( 1 - m ) x = 3 ( 1 + m ) :‬‬‫* ﻤﻥ ﺃﺠل ‪ 1 - m = 0‬ﺃﻱ ‪ m = 1‬ﻟﺩﻴﻨﺎ ‪ 0 × x = 6 :‬ﻭﻫﺫﺍ ﻤﺴﺘﺤﻴل ‪.‬‬‫‪x‬‬ ‫=‬ ‫)‪3(1+ m‬‬ ‫* ﻤﻥ ﺃﺠل ‪ 1 - m ≠ 0‬ﺃﻱ ‪ m ≠ 1‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪1-m‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m = 1‬ﻟﺩﻴﻨﺎ ‪x = 3 :‬‬ ‫‪.‬‬ ‫‪x=3‬‬ ‫ﺃﻭ‬ ‫=‪x‬‬ ‫)‪3(1+ m‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m ≠ 1‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪1-m‬‬ ‫‪(3) . . . ( x – 5 ) ( x 2 – 6x + 9 ) ( x 2 + m x – m 2 ) = 0 ( 3‬‬ ‫ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪ x - 5 = 0‬ﺃﻭ ‪ x2 - 6x + 9 = 0‬ﺃﻭ‬ ‫‪x2 + m x - m2 = 0‬‬ ‫* ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x - 5 = 0 :‬ﺘﻌﻨﻲ ‪x = 5 :‬‬ ‫* ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x2 - 6x + 9 = 0 :‬ﻭﻤﻨﻪ ‪∆ = b2 - 4ac :‬‬

‫ﻭﻋﻠﻴﻪ ‪ ∆ = 36 - 36 = 0 :‬ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ ﻫﻭ ‪:‬‬‫‪x0‬‬ ‫=‬ ‫‪-b‬‬ ‫=‬ ‫‪6‬‬ ‫‪=3‬‬ ‫‪2a‬‬ ‫‪2‬‬ ‫* ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ x2 + m x - m2 = 0 :‬ﻨﺤﺴﺏ ﺍﻟﻤﻤﻴﺯ ‪∆ = b2 - 4ac‬‬ ‫ﻭﻋﻠﻴﻪ ‪ ∆ = m2 - 4 ( 1 ) - m2 :‬ﺃﻱ ‪( )∆ = 5m2 :‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m = 0‬ﻓﺈﻥ ‪ ∆ = 0 :‬ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋـﻑ ‪:‬‬‫‪x0‬‬ ‫=‬ ‫‪-m‬‬ ‫‪=0‬‬ ‫‪2‬‬ ‫* ﻤﻥ ﺃﺠل ‪ m ≠ 0‬ﻓﺈﻥ ‪∆ > 0 :‬‬ ‫ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪:‬‬ ‫= ‪x1‬‬ ‫‪-b-‬‬ ‫∆‬ ‫=‬ ‫= ‪- m - 5m2‬‬ ‫‪-m-‬‬ ‫‪m‬‬ ‫‪5‬‬ ‫‪2a‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪x2‬‬ ‫=‬ ‫‪-b+‬‬ ‫=∆‬ ‫= ‪- m + 5m2‬‬ ‫‪-m+‬‬ ‫‪m‬‬ ‫‪5‬‬ ‫ﻭ‬ ‫‪2a‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﻭ ﻤﻨﻪ ﻨﺨﻠﺹ ﺇﻟﻰ ‪:‬‬ ‫* ﻟﻤﺎ ‪ : m = 0‬ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ) ‪ ( 3‬ﻫﻲ ‪{ }E = 5 , 3 , 0 :‬‬ ‫ﺤﻴﺙ ‪ 3 , 0 :‬ﻫﻤﺎ ﺤﻠﻴﻥ ﻤﻀﺎﻋﻔﻴﻥ ‪.‬‬ ‫* ﻟﻤﺎ ‪ : m ∈ - 0‬ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ) ‪ ( 3‬ﻫﻲ ‪{ }:‬‬‫‪E‬‬ ‫=‬ ‫‪‬‬ ‫‪5,3,‬‬ ‫‪-m+ a‬‬ ‫‪5‬‬ ‫‪,‬‬ ‫‪-m- a‬‬ ‫‪5 ‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪ 3‬ﺤل ﻤﻀﺎﻋﻑ ‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 5‬‬‫ﻨﻔﺭﺽ ‪ x‬ﻁﻭل ﺍﻟﻘﺎﻋﺩﺓ ﺍﻟﺼﻐﺭﻯ ﻓﻴﻜﻭﻥ ‪ 2 x‬ﻁﻭل ﻜل ﻤﻥ ﺍﻟﻘﺎﻋﺩﺓ ﺍﻟﻜﺒﺭﻯ ﻭ ﺍﻻﺭﺘﻔﺎﻉ ﻤﺴﺎﺤﺔ ﺍﻟﻘﻁﻌﺔ‬ ‫‪S‬‬ ‫=‬ ‫)‪(2x + x‬‬ ‫ﺍﻷﺜﺭﻴﺔ ﻫﻲ ‪× 2 x :‬‬ ‫‪2‬‬ ‫‪x2 = 48‬‬ ‫‪3‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﻭﻤﻨﻪ ‪3 x2 = 48 :‬‬ ‫‪S = 3x2‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬

‫ﺃﻱ ‪ x2 = 16 :‬ﻭﻋﻠﻴﻪ ‪. x = 4 :‬‬ ‫ﻭﻋﻠﻴﻪ ﻁﻭل ﻗﺎﻋﺩﺘﻪ ﺍﻟﺼﻐﺭﻯ ‪ 4 Cm‬ﻭﻁﻭل ﻗﺎﻋﺩﺘﻪ ﺍﻟﻜﺒﺭﻯ ‪8 Cm‬‬ ‫ﻭﻁﻭل ﺍﺭﺘﻔﺎﻋﻪ ‪. 8 Cm‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 6‬‬ ‫ﻤﺴﺎﺤﺔ ‪ ABCD‬ﻫﻲ ‪S1 = ( 2x - 1) x = 2x2 - x :‬‬ ‫‪S2‬‬ ‫=‬ ‫‪(2x - 1) x‬‬ ‫=‬ ‫‪2 x2 - x‬‬ ‫ﻤﺴﺎﺤﺔ ‪ ADE‬ﻫﻲ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪S = s1 + s2‬‬ ‫ﻤﺴﺎﺤﺔ ﺍﻟﻤﻨﺯل ﻫﻲ‪:‬‬‫‪S = 2 x2‬‬ ‫‪-x+‬‬ ‫‪2 x2 - x‬‬ ‫=‬ ‫‪4 x2‬‬ ‫‪- 2 x + 2 x2‬‬ ‫‪-x‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪S = 6 x2 - 3x‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫‪6‬‬ ‫‪x2 -‬‬ ‫‪3‬‬ ‫‪x‬‬ ‫‪= 180‬‬ ‫ﻭﻟﺩﻴﻨﺎ ‪ S = 180 m2 :‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪2‬‬‫ﺇﺫﻥ ‪ 6 x2 - 3 x = 360 :‬ﺃﻱ ‪6 x2 - 3 x - 360 = 0 :‬‬ ‫ﺒﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ ‪ 3‬ﻨﺠﺩ ‪2 x2 - x - 120 = 0 :‬‬‫ﻟﺩﻴﻨﺎ ‪ ∆ = b2 - 4ac :‬ﺃﻱ ‪∆ = (-1)2 - 4(2)(-120) = 961 :‬‬ ‫‪ ∆ > 0‬ﻭ ﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪:‬‬‫‪x1‬‬ ‫=‬ ‫‪-b-‬‬ ‫∆‬ ‫=‬ ‫‪1‬‬ ‫‪- 961‬‬ ‫=‬ ‫‪−15‬‬ ‫) ﻤﺭﻓﻭﺽ (‬ ‫‪2a‬‬ ‫‪2×2‬‬ ‫‪2‬‬‫‪x2‬‬ ‫=‬ ‫‪-b+‬‬ ‫∆‬ ‫=‬ ‫‪1‬‬ ‫‪+ 961‬‬ ‫=‬ ‫‪8‬‬ ‫ﻭ‬ ‫‪2a‬‬ ‫‪2×2‬‬ ‫ﻭﻋﻠﻴﻪ ‪. x = 8 :‬‬

‫ﺘﻤﺭﻴﻥ ‪: 7‬‬ ‫ﻨﻔﺭﺽ ‪ x‬ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ؛ ﻓﻴﻜﻭﻥ ﺴﻌﺭ ﺍﻟﻜﺘﺎﺏ ‪. x2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ 10 x + 5 x2 = 2200 :‬ﺃﻱ ‪5 x2 + 10 x - 2200 = 0 :‬‬ ‫ﺒﺎﻟﻘﺴﻤﺔ ﻋﻠﻰ ‪ 5‬ﻨﺠﺩ ‪ x2 + 2 x - 440 = 0 :‬ﻭﻫﻲ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ‪ .‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ ∆ = b2 - 4ac‬ﺇﺫﻥ ‪∆ = 22 - 4 (1) (-440) = 1764 :‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪ x1 , x2‬ﺤﻴﺙ ‪:‬‬ ‫) ﻤﺭﻓﻭﺽ (‬ ‫= ‪x1‬‬ ‫‪-b-‬‬ ‫∆‬ ‫=‬ ‫‪- 2 - 1764‬‬ ‫=‬ ‫‪- 2 - 42‬‬ ‫‪= - 22‬‬ ‫‪2a‬‬ ‫)‪2(1‬‬ ‫‪2‬‬ ‫َو‬ ‫‪x2‬‬ ‫=‬ ‫‪-b‬‬ ‫‪+‬‬ ‫∆‬ ‫=‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪+ 1764‬‬ ‫=‬ ‫‪- 2 + 42‬‬ ‫=‬ ‫‪20‬‬ ‫‪2a‬‬ ‫)‪2(1‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪ x = 20 :‬ﻭﻋﻠﻴﻪ ‪x2 = 400 :‬‬ ‫ﺇﺫﻥ ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ﺍﻟﻭﺍﺤﺩ ﻫﻭ ‪ 20 DA :‬ﻭﺴﻌﺭ ﺍﻟﻜﺘﺎﺏ ﺍﻟﻭﺍﺤﺩ ﻫﻭ ‪. 400 DA :‬‬ ‫ﺘﻤﺭﻴﻥ ‪A : 8‬‬ ‫ﻨﻔﺭﺽ ‪ x‬ﻁﻭل ‪ AC‬ﻓﻴﻜﻭﻥ ‪[ ]:‬‬ ‫‪x‬‬ ‫‪ x + 1‬ﻁﻭل ‪[ ]. BC‬‬ ‫‪4 Cm‬‬ ‫‪HC‬‬ ‫=‬ ‫‪1‬‬ ‫‪BC‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪HC‬‬ ‫=‬ ‫‪2‬‬ ‫)‪(x+1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫‪B‬‬ ‫‪x +1‬‬ ‫‪H‬‬ ‫‪C‬‬ ‫ﻓﻲ ﺍﻟﻤﺜـﻠﺙ ﺍﻟﻘﺎﺌﻡ ‪ AHC‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2x2‬‬ ‫=‬ ‫‪ x+12‬‬ ‫‪+ 42‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪AC2 = AH2 + HC2‬‬ ‫‪ 2 ‬‬

‫ﺃﻱ ‪3 x2 - 2x - 65 = 0 :‬‬ ‫‪x2 = x2 + 2x +1‬‬ ‫‪+ 16‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪4‬‬‫ﻟﺩﻴﻨﺎ ‪ ∆ = b2 - 4ac :‬ﺃﻱ ‪∆ = (-2)2 - 4(3)(-65) = 784 :‬‬ ‫ﺇﺫﻥ ‪ ∆ > 0 :‬ﻭﻋﻠﻴﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬‫= ‪x1‬‬ ‫‪-b -‬‬ ‫∆‬ ‫=‬ ‫‪2 - 784‬‬ ‫=‬ ‫‪2 - 28‬‬ ‫=‬ ‫‪- 13‬‬ ‫) ﻤﺭﻓﻭﺽ (‬ ‫‪2a‬‬ ‫)‪2(3‬‬ ‫‪6‬‬ ‫‪2‬‬‫‪x2‬‬ ‫=‬ ‫‪-b +‬‬ ‫∆‬ ‫=‬ ‫‪2 + 784‬‬ ‫=‬ ‫‪2 + 28‬‬ ‫‪=5‬‬ ‫َﻭ‬ ‫‪2a‬‬ ‫)‪2(3‬‬ ‫‪6‬‬ ‫ﺇﺫﻥ ‪. x = 5 :‬‬‫ﻭﻋﻠﻴﻪ ﻁﻭل ﻜل ﻤﻥ ‪ AB‬و ‪ AC‬ﻫﻭ ‪َ 5 Cm :‬ﻭ ﻁﻭل ‪ BC‬ﻫﻭ‪[ ] [ ] [ ]6 Cm :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 9‬‬‫ﺒﻤﺎ ﺃﻥ ﺍﻟﻭﺘﺭ ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ ﻫﻭ ﺃﻁﻭل ﻀﻠﻊ ﻓﺈﻨﻨﺎ ﻨﻔﺭﺽ ﺃﻥ ﻁﻭل ﺍﻟﻭﺘﺭ ﻫﻭ ‪ x + 2‬ﻭ ﻤﻨﻪ‬ ‫ﻁﻭﻟﻲ ﺍﻟﻀﻠﻌﻴﻥ ﺍﻟﻘﺎﺌﻤﻴﻥ ﻫﻤﺎ ‪x ; x + 1 :‬‬ ‫ﻭﺤﺴﺏ ﻨﻅﺭﻴﺔ ﻓﻴﺘﺎﻏﻭﺭﺙ ﻴﻨﺘﺞ ‪( x + 2 )2 = ( x + 1 )2 + x2 :‬‬ ‫ﻭﻤﻨﻪ ‪x2 + 4 x + 4 = x2 + 2 x + 1 + x2 :‬‬ ‫ﺃﻱ ‪ x2 - 2 x - 3 = 0 :‬ﻭ ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻓﻴﻬﺎ ‪a + c = b :‬‬ ‫ﺇﺫﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻫﻤﺎ‪َ - 1 :‬ﻭ ‪ 3‬ﻟﻜﻥ ﺍﻟﺤل ‪ - 1‬ﻤﺭﻓﻭﺽ ﺇﺫﻥ ‪. x = 3 :‬‬ ‫ﻭﻋﻠﻴﻪ ﻁﻭل ﺍﻟﻭﺘﺭﻫﻭ ‪ 5 Cm‬ﻭﻁﻭﻟﻲ ﺍﻟﻀﻠﻌﻴﻥ ﺍﻟﻘﺎﺌﻤﻴﻥ ﻫﻤﺎ ‪4 Cm , 3 Cm :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 10‬‬‫)‪15 - x = 8 (8 - x‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪15 - x = 8‬‬ ‫ﻨﺴﻤﻲ ‪ x‬ﻫﺫﺍ ﺍﻟﻌﺩﺩ ﻓﻨﺠﺩ ‪:‬‬ ‫‪8-x‬‬ ‫ﺃﻱ ‪ 15 - x = 64 - 8x :‬ﻭﻤﻨﻪ ‪−x + 8x = 64 - 15 :‬‬ ‫=‪x‬‬ ‫‪49‬‬ ‫‪=7‬‬ ‫ﺃﻱ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ‪7x = 49 :‬‬ ‫‪7‬‬ ‫ﺇﺫﻥ ﺍﻟﻌﺩﺩ ﺍﻟﻤﻁﻠﻭﺏ ﻫﻭ ‪. x = 7 :‬‬

‫ﺘﻤﺭﻴﻥ ‪: 11‬‬‫ﻟﻴﻜﻥ ‪ x‬ﻋﺩﺩ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ﻓﻲ ﻜل ﺼﻑ ﻤﻥ ﺃﻀﻼﻉ ﺍﻟﻤﺭﺒﻊ ﺍﻷﻭل ﻓﻴﻜﻭﻥ ﻋﺩﺩ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ‪x 2 + :‬‬ ‫‪ 11‬ﻭﻋﻠﻴﻪ ﻴﻜﻭﻥ ﻋﺩﺩ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ﺍﻟﺘﻲ ﺍﺴﺘﻌﻤﻠﻬﺎ ﻋﻠﻰ ﻁﻭل ﻀﻠﻊ ﺍﻟﻤﺭﺒﻊ ﺍﻟﺜﺎﻨﻲ ﻫﻭ ‪x + 1‬‬ ‫ﻭﻤﻨﻪ ﻋﺩﺩ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ﻫﻭ ‪. ( x + 1 ) 2 – 16 :‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪x 2 + 11 = ( x + 1 ) 2 - 16 :‬‬‫ﻭﻤﻨﻪ ‪ x 2 + 11 – x 2 – 2x – 1 + 16 = 0 :‬ﺃﻱ ‪- 2x + 26 = 0 :‬‬ ‫ﻭﻤﻨﻪ ‪. x = 13 :‬‬‫ﻋﺩﺩ ﺯﻫﺭﺍﺕ ﺍﻟﻨﺭﺩ ‪ ( 13 ) 2 + 11 :‬ﺇﺫﻥ ﻋﺩﺩﻫﺎ ﻫﻭ ‪ 180 :‬ﺯﻫﺭﺓ ﻨﺭﺩ ‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 12‬‬ ‫ﻨﻔﺭﺽ ﻫﺫﻩ ﺍﻷﻋﺩﺍﺩ ﻫﻲ ‪x ; x + 1 ; x + 2 :‬‬ ‫ﻟﺩﻴﻨﺎ ‪( x + 1 ) ( x + 2 ) – x 2 = 41 :‬‬‫ﺃﻱ ‪ x 2 + 3x + 2 – x 2 = 41 :‬ﻭﻋﻠﻴﻪ ‪3x = 39 :‬‬‫ﻭﻋﻠﻴﻪ ﺍﻷﻋﺩﺍﺩ ﻫﻲ ‪13 , 14 , 15 :‬‬ ‫ﺇﺫﻥ ‪. x = 13 :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 13‬‬ ‫ﺘﻌﻴﻴﻥ ﻭﻀﻌﻴﺔ ‪: M‬‬‫* ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ ‪ ABM‬ﻟﺩﻴﻨﺎ ‪AM 2 = AB 2 + BM 2 :‬‬ ‫ﻭﻤﻨﻪ ‪AM 2 = 9 + ( BC – MC ) 2 :‬‬ ‫ﺇﺫﻥ ‪AM 2 = 9 + ( 5 – MC ) 2 :‬‬‫* ﻭ ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ ‪ DCM‬ﻟﺩﻴﻨﺎ ‪DM 2 = DC 2 + MC 2 :‬‬ ‫ﻭﻤﻨﻪ ‪DM 2 = 16 + MC 2 :‬‬‫ﻴﻜﻭﻥ ‪ AM = DM‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪AM 2 = DM 2 :‬‬ ‫ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ‪9 + ( 5 – MC ) 2 = 16 + MC 2 :‬‬ ‫ﺃﻱ ‪9 + 25 – 10MC + MC 2 = 16 + MC 2 :‬‬‫‪. MC = 1,8 Cm‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪MC‬‬ ‫=‬ ‫‪18‬‬ ‫‪:‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪10‬‬‫ﻭﻋﻠﻴﻪ ‪ MB = 5 – 1,8 :‬ﺃﻱ ‪. MB = 3,2 Cm :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 14‬‬‫‪x ≥ 20‬‬ ‫ﻨﻔﺭﺽ ‪ x‬ﺴﻌﺭ ﺍﻟﻘﻠﻡ ‪ .‬ﻓﻴﻜﻭﻥ ‪ 2x‬ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ‪.‬‬ ‫ﻟﺩﻴﻨﺎ ‪َ 8 x + 20 (2x) < 1000 :‬ﻭ‬

‫‪x‬‬ ‫<‬ ‫‪1000‬‬ ‫ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺍﻷﻭﻟﻰ ﺘﻌﻨﻲ ﺃﻥ ‪ 48 x < 1000 :‬ﺃﻱ ‪:‬‬ ‫‪48‬‬ ‫ﺇﺫﻥ ‪ x < 20,83 :‬ﻭ ﺒﻤﺎ ﺃﻥ ‪ x ≥ 20 :‬ﻓﺈﻥ ‪:‬‬ ‫‪. x = 20‬‬ ‫ﻭ ﻋﻠﻴﻪ ‪ :‬ﺴﻌﺭ ﺍﻟﻘﻠﻡ ﻫﻭ ‪َ 20 DA‬ﻭ ﺴﻌﺭ ﺍﻟﻜﺭﺍﺱ ﻫﻭ ‪. 40 DA‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 15‬‬ ‫‪−x + 3‬‬ ‫≤‬ ‫‪ ( 1‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪0 :‬‬ ‫‪2x + 5‬‬ ‫‪−x + 3‬‬ ‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ‪: 2x + 5‬‬‫‪x‬‬ ‫∞‪−‬‬ ‫‪5‬‬ ‫∞‪3 +‬‬ ‫‪-2‬‬‫‪-x+3‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫ــ‬‫‪2x + 5‬‬ ‫ــ‬ ‫‪+‬‬ ‫‪+‬‬‫‪−x + 3‬‬ ‫ــ‬ ‫ــ‬‫‪2x + 5‬‬‫‪‬‬ ‫‪−∞ ,‬‬ ‫‪-5‬‬ ‫‪‬‬ ‫‪U  3 , + ∞ ‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪:‬‬‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪ ( 2‬ﺤل ﺍﻟﺘﺭﺍﺠﺤﺔ ‪(− 5x + 2) (4x + 5) > 0 :‬‬ ‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ‪(− 5x + 2)( 4 x + 5 ) :‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪−5‬‬ ‫‪2‬‬ ‫∞‪+‬‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪4x+5‬‬‫‪-5x + 2‬‬ ‫‪ + +‬ــ‬ ‫اﻟﺠﺪاء‬ ‫ــ ‪+ +‬‬ ‫ــ ‪ +‬ــ‬

‫‪‬‬ ‫‪-5‬‬ ‫‪,‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪:‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪5 ‬‬ ‫‪2x‬‬ ‫‪-‬‬ ‫‪2x + 4‬‬ ‫‪ ( 3‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪≥ 0 :‬‬ ‫‪x +4‬‬ ‫‪x-5‬‬ ‫‪2x‬‬ ‫)‪(x - 5) - (x + 4) (2 x + 4‬‬ ‫≥‬ ‫‪0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫)‪(x + 4) (x - 5‬‬ ‫‪2 x2 - 10 x - 2 x2 - 4x - 8x -16‬‬ ‫≥‬ ‫‪0‬‬ ‫ﺃﻱ ‪:‬‬ ‫)‪(x + 4) (x - 5‬‬ ‫‪− 22 x - 16‬‬ ‫‪≥0‬‬ ‫ﺃﻱ ‪:‬‬ ‫)‪(x + 4) (x - 5‬‬ ‫‪− 22 x - 16‬‬ ‫ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ‪( x + 4)( x - 5) :‬‬ ‫‪x‬‬ ‫‪−∞ -4‬‬ ‫‪8‬‬ ‫∞‪5 +‬‬ ‫‪- 11‬‬ ‫‪- 22 x - 16‬‬ ‫‪++‬‬ ‫ــ‬ ‫‪x+4‬‬ ‫‪ +‬ــ‬ ‫ــ‬ ‫‪+‬‬ ‫‪x-5‬‬ ‫ــ ــ‬ ‫‪+‬‬ ‫‪+‬‬ ‫ــ ‪+‬‬ ‫ــ‬ ‫‪+‬‬ ‫)‪(x+4)(x-5‬‬ ‫ــ‬ ‫‪− 22 x - 16‬‬‫)‪(x + 4)(x - 5‬‬‫‪ - ∞ , - 4  U‬‬ ‫‪ -8‬‬ ‫‪,‬‬ ‫‪5‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪:‬‬ ‫‪ 11‬‬ ‫‪‬‬‫‪ ( 4‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪4 x2 − 25 - ( 2 x - 5 ) (x + 3) ≥ 0 :‬‬

‫ﺃﻱ ‪(2x - 5 )(2x + 5) - (2x - 5)(x + 3) ≥ 0 :‬‬ ‫‪(2x - 5) (2x + 5) -(x + 3) ≥ 0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪(2x - 5) (x + 2) ≥ 0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ‪(2x - 5) ( x + 2) :‬‬ ‫‪x‬‬ ‫‪−∞ -2‬‬ ‫‪5‬‬ ‫∞‪+‬‬ ‫‪2‬‬‫‪ +‬ــ ــ ‪2 x - 5‬‬‫‪ + +‬ــ ‪x + 2‬‬‫)‪(2x - 5) (x + 2‬‬ ‫‪+‬‬ ‫ــ‬ ‫‪+‬‬ ‫‪‬‬ ‫∞‪-‬‬ ‫‪,-2‬‬ ‫‪‬‬ ‫‪U‬‬ ‫‪‬‬ ‫‪5‬‬ ‫‪,‬‬ ‫‪+‬‬ ‫∞‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪ ( 5‬ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪20 x2 + 6x - 2 < 0 :‬‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = 62 − 4 × 20 × (-2) = 196 :‬‬‫= ‪x1‬‬ ‫‪-b -‬‬ ‫∆‬ ‫=‬ ‫‪- 6 - 196‬‬ ‫=‬ ‫‪- 6 - 14‬‬ ‫=‬ ‫‪-1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪2a‬‬ ‫)‪2(20‬‬ ‫‪40‬‬ ‫‪2‬‬‫‪x2‬‬ ‫=‬ ‫‪-b -‬‬ ‫∆‬ ‫=‬ ‫‪- 6 + 196‬‬ ‫=‬ ‫‪- 6 + 14‬‬ ‫=‬ ‫‪1‬‬ ‫َو‬ ‫‪2a‬‬ ‫)‪2(20‬‬ ‫‪40‬‬ ‫‪5‬‬ ‫ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ‪20 x2 + 6x - 2 :‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪-1‬‬ ‫‪1‬‬ ‫∞‪+‬‬ ‫‪2‬‬ ‫‪5‬‬‫‪20 x2 + 6x - 2‬‬ ‫‪+‬‬ ‫ــ‬ ‫‪+‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪1‬‬ ‫‪‬‬ ‫ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل‬ ‫‪‬‬ ‫‪2‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪ ( 6‬ﺤـل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪49 x2 - 14 x + 1 ≤ 0 :‬‬

‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = ( - 14 )2 - 4 ( 49 )(1) = 0 :‬‬ ‫‪x0‬‬ ‫=‬ ‫‪-b‬‬ ‫=‬ ‫‪14‬‬ ‫=‬ ‫‪1‬‬ ‫ﻭﻤﻨﻪ ﺍﻟﺠﺫﺭ ﺍﻟﻤﻀﺎﻋﻑ ﻫﻭ ‪:‬‬ ‫‪2a‬‬ ‫‪2 × 49‬‬ ‫‪7‬‬ ‫‪49‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪1 2‬‬ ‫≤‬ ‫‪0‬‬ ‫ﺇﺫﻥ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺘﺼﺒﺢ ‪:‬‬ ‫‪‬‬ ‫‪7 ‬‬‫‪.‬‬ ‫=‪x‬‬ ‫‪1‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪x−‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫ﻭﻫﺫﺍ ﺨﺎﻁﺊ ﺇﻻ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 16‬‬ ‫‪ ( 1‬ﺍﻟﺤل ﺍﻟﺠﺒﺭﻱ ‪:‬‬ ‫ﻨﻔﺭﺽ ‪ x‬ﻋﺭﺽ ﺍﻟﻤﺴﺘﻁﻴل ﻭ ‪ y‬ﻁﻭﻟﻪ ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪. x < y‬‬ ‫َﻭ ‪2 ( x + y ) = 10‬‬ ‫ﻓﻴﻜﻭﻥ ‪x y = 6 :‬‬ ‫‪x+y =5‬‬ ‫َﻭ‬ ‫ﺇﺫﻥ ‪x y = 6 :‬‬ ‫‪y= 5–x‬‬ ‫َﻭ‬ ‫ﻭﻋﻠﻴﻪ ‪x y = 6 :‬‬ ‫ﻭﻤﻨﻪ ‪َ x ( 5 - x ) = 6 :‬ﻭ‬ ‫‪y=5–x‬‬ ‫ﺇﺫﻥ ‪ − x2 + 5 x - 6 = 0 :‬ﻭ ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = 52 - 4 (-1)(-6) = 1 :‬‬‫= ‪x1‬‬ ‫‪-b -‬‬ ‫∆‬ ‫=‬ ‫‪- 5 -1‬‬ ‫‪=3‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻫﻤﺎ ‪:‬‬ ‫‪2a‬‬ ‫)‪2 (-1‬‬ ‫‪x2‬‬ ‫=‬ ‫‪-b +‬‬ ‫∆‬ ‫=‬ ‫‪- 5 +1‬‬ ‫‪=2‬‬ ‫َو‬ ‫‪2a‬‬ ‫)‪2 (-1‬‬ ‫‪ ) y = 2‬ﻤﺭﻓﻭﺽ ﻷﻥ ‪( x < y‬‬ ‫ﻟﻤﺎ ‪ x = 3‬ﻓﺎﻥ ‪:‬‬ ‫‪y= 3‬‬ ‫ﻟﻤﺎ ‪ x = 2‬ﻓﺎﻥ ‪:‬‬ ‫ﻭﻤﻨﻪ ﻴﻭﺠﺩ ﻤﺴﺘﻁﻴل ﺤﻴﺙ ﻁﻭﻟﻪ ‪َ 3 Cm‬ﻭ ﻋﺭﻀﻪ ‪. 2 Cm‬‬ ‫‪x +y=5‬‬ ‫و‬ ‫‪ ( 2‬ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ‪:‬‬ ‫‪y=5 - x‬‬ ‫َﻭ‬ ‫ﻟﺩﻴﻨﺎ ‪x y = 6 :‬‬ ‫ﺤﻴﺙ ‪x < y :‬‬ ‫‪y‬‬ ‫=‬ ‫‪6‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪x‬‬

‫‪g (x) = 5 - x‬‬ ‫َﻭ‬ ‫= )‪f (x‬‬ ‫‪6‬‬ ‫ﻟﺘﻜﻥ ‪ g , f‬ﺍﻟﺩﺍﻟﺘﻴﻥ ﺍﻟﻤﻌﺭﻓﺘﻴﻥ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪x‬‬‫ﻨﻘﻭﻡ ﺒﺈﻨﺸﺎﺀ ﺍﻟﺒﻴﺎﻨﻴﻥ )‪ (δ‬و )‪ (c‬ﺍﻟﻤﻤﺜﻠﻴﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ ‪ g , f‬ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺤﺎﺴﺒﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ﺃﻭ ﺍﻟﺤﺎﺴﻭﺏ‬ ‫ﺒﺘﺒﺎﻉ ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪ / 1‬ﻨﺴﺘﻌﻤل ﺍﻟﻠﻤﺴﺔ )‪ (1‬ﻟﺘﺤﻭﻴل ﺍﻟﻨﺹ ﻤﻥ ﺁﺩﺒﻲ ‪ T‬ﺇﻟﻰ ﺭﻴﺎﻀﻲ ‪. M‬‬ ‫‪ / 2‬ﻨﻜﺘﺏ ﻋﺒﺎﺭﺓ ﺍﻟﺩﺍﻟﺘﻴﻥ ﻓﻲ ﻟﻭﺤﺔ ﺘﺤﺭﻴﺭ ﺍﻟﻨﺹ )‪. (2‬‬ ‫‪ / 3‬ﻨﺴﺘﻌﻤل ﺍﻟﻠﻤﺴﺔ )‪ (3‬ﻟﺘﻤﺜﻴل ﺍﻟﺩﺍﻟﺘﻴﻥ ﻓﻲ ﻤﻌﻠﻡ ﺜﻨﺎﺌﻲ ﺍﻟﺒﻌﺩﻴﻥ ‪.‬‬ ‫ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺴﺎﺒﻘﺔ ﻤﻭﻀﺤﺔ ﻓﻲ ﺍﻟﺭﺴﻡ ﺍﻟﺘﺎﻟﻲ ‪:‬‬ ‫)‪(1‬‬ ‫)‪(3‬‬ ‫)‪(2‬‬ ‫‪ / 4‬ﻨﻀﺒﻁ ﺍﻟﻭﺤﺩﺍﺕ ﻭ ﻗﻴﻡ ‪ x‬ﻭ ﻜﺫﻟﻙ ﻗﻴﻡ ‪ y‬ﻭ ﻫﺫﺍ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻠﻤﺴﺔ )‪. (4‬‬‫‪ / 5‬ﻨﺴﺘﻌﻤل ﺍﻟﻠﻤﺴﺔ )‪ (5‬ﻟﻜﻲ ﻨﻅﻬﺭ ﻟﻭﺤﺔ )‪ (6‬ﺍﻟﺘﻲ ﺘﻌﻁﻲ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺃﻱ ﻨﻘﻁﺔ ﻤﻥ‬ ‫ﺍﻟﻤﻌﻠﻡ ﺤﻴﺙ ‪ Horz‬ﻴﻤﺜل ‪ x‬ﻭ ‪ Vert‬ﻴﻤﺜل ‪. y‬‬ ‫ﺍﻟﺨﻁﻭﺘﺎﻥ ﺍﻷﺨﻴﺭﺘﺎﻥ ﺘﻅﻬﺭﺍﻥ ﻓﻲ ﺍﻟﺭﺴﻡ ﺍﻟﺘﺎﻟﻲ ‪:‬‬

‫)‪(δ‬‬ ‫)‪(5‬‬ ‫)‪(c‬‬ ‫)‪(6‬‬‫)‪(δ‬‬ ‫)‪(4‬‬‫ﻨﻘﻁﺘﻲ ﺘﻘﺎﻁﻊ )‪ (δ‬و )‪ (c‬ﻫﻤﺎ ﺍﻟﻨﻘﻁﺘﺎﻥ ‪ M1( 3 , 2 ) :‬ﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻬﺎ ﻓﻲ ﺍﻟﺸﻜل ﺒﺎﻟﺴﻬﻡ ‪:‬‬‫َﻭ ﻫﻲ ﺍﻟﻨﻘﻁﺔ ﺍﻟﺘﻲ ﺘﻅﻬﺭ ﺇﺤﺩﺍﺜﻴﺎﺘﻬﺎ‬ ‫َﻭ ) ‪ M2 ( 2 , 3‬ﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻬﺎ ﻓﻲ ﺍﻟﺸﻜل ﺒﺎﻟﺴﻬﻡ ‪:‬‬ ‫ﺒﺎﻟﺘﻘﺭﻴﺏ ﻓﻲ ﺍﻟﻠﻭﺤﺔ )‪(6‬‬ ‫ﺇﺫﻥ ‪y ≅ 2,9727 , x ≅ 2 ,0284 :‬‬ ‫ﻭﻤﻨﻪ ‪ x = 2 :‬و ‪y = 3‬‬ ‫ﺇﺫﻥ ﻴﻭﺠﺩ ﻤﺴﺘﻁﻴل ﺤﻴﺙ ﻁﻭﻟﻪ ‪َ 3 Cm‬ﻭ ﻋﺭﻀﻪ ‪. 2 Cm‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 17‬‬ ‫‪ (1‬ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪: f (x) ≤ 0‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ ‪[-3 , -1] U [2 , 4] :‬‬ ‫ﻭﻫﻲ ﻓﻭﺍﺼل ﺍﻟﻨﻘﻁ ﺍﻟﺘﻲ ﺘﻘﻊ ﺘﺤﺕ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ‪.‬‬ ‫‪ (2‬ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﻤﻌﺎﺩﻟﺔ ‪. f (x) = -1‬‬‫ﻨﺭﺴﻡ ﺍﻟﻤﺴﺘﻘﻴﻡ ‪ ∆1‬ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ y = -1‬ﻓﻴﻘﻁﻊ ‪ γ‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ﺍﻟﺘﻲ ﻓﻭﺍﺼﻠﻬﺎ ﺘﻨﺘﻤﻲ ﺇﻟﻰ) ( ) (‬ ‫ﺍﻟﻤﺠﻤﻭﻋﺔ ‪ -3 , -2 ∪ 3 :‬ﻭﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪[ ] { }.‬‬ ‫‪ (3‬ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﻤﻌﺎﺩﻟﺔ ‪: f (x) = - x + 1‬‬

‫ﻨﺭﺴﻡ ﺍﻟﻤﺴﺘﻘﻴﻡ ‪ ∆2‬ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪ y = -x + 1 :‬ﻓﻴﻘﻁﻊ ‪ γ‬ﻓﻲ ﺍﻟﻨﻘﻁﺘﻴﻥ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺘﻴﻥ) ( ) (‬ ‫‪َ 0‬ﻭ ‪ 3‬ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪{ }0 , 3 :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 18‬‬ ‫* ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﻤﻌﺎﺩﻟﺔ ‪: 4 - x2 = - 5‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪f (x) = 4 - x2 :‬‬ ‫ﻨﻨﺸﺊ ﺒﻭﺍﺴﻁﺔ ﺍﻟﺤﺎﺴﺒﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ﺃﻭ ﺍﻟﺤﺎﺴﻭﺏ ﺍﻟﻤﻨﺤﻨﻰ ‪ δ‬ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻓﻨﺠﺩ ‪( ):‬‬ ‫)‪(δ‬‬ ‫‪3‬‬‫‪-3‬‬ ‫‪M1‬‬ ‫‪M2‬‬‫)∆(‬ ‫ﺜﻡ ﻨﻨﺸﺊ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪ y = -5 :‬ﻓﻴﻘﻁﻊ ‪ δ‬ﻓﻲ ﺍﻟﻨﻘﻁﺘﻴﻥ ‪( ) ( ):‬‬ ‫) ‪َ M2 ( 3 , -5‬ﻭ ) ‪. M1( -3 , -5‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ ‪{ }. -3 , 3 :‬‬‫* ﺍﻟﺤﻠﻭل ﺍﻟﺒﻴﺎﻨﻴﺔ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪ 4 - x2 ≥ 0‬ﻫﻲ ﻓﻭﺍﺼل ﺍﻟﻨﻘﻁ ﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻓﻴﻬﺎ ‪ δ‬ﻓﻭﻕ) (‬ ‫ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﺃﻱ ‪f (x) ≥ 0 :‬‬‫ﻓﻲ ﺍﻟﻤﺠﺎل ‪ δ : −2 , 2‬ﻴﻘﻁﻊ ﻓﻭﻕ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪( ) [ ]:‬‬ ‫]‪. [-2 , 2‬‬

‫* ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺠﺒﺭﻴﺔ ‪:‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪4 - x2 = -5 :‬‬ ‫ﺃﻱ ‪ x2 = 9 :‬ﻭ ﻋﻠﻴﻪ ‪ x = 3 :‬ﺃﻭ ‪x = -3‬‬ ‫ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪: 4 - x2 ≥ 0 :‬‬ ‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ‪ 4 - x2‬ﺃﻱ ‪( 2 - x )( 2 + x ) :‬‬‫ﻭﺨﺎﺭﺝ ﺍﻟﺠﺫﺭﻴﻥ‬ ‫ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻟﻪ ﺠﺫﺭﻴﻥ ﻫﻤﺎ ‪َ 2‬ﻭ ‪ - 2‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﺇﺸﺎﺭﺘﻪ ﺒﻴﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺴﺎﻟﺒﺔ‬ ‫ﻤﻭﺠﺒﺔ ﻭﻋﻠﻴﻪ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪[ ]-2 , 2 :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 19‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ﻨﻨﺸﺊ ﺍﻟﺘﻤﺜﻴﻼﺕ ﺍﻟﺒﻴﺎﻨﻴﺔ ‪( ) ( ) ( )U1 ; U2 ; U3‬‬ ‫ﻟﻠﺩﻭﺍل ‪ f , g , h‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﺤﻴﺙ ‪f ( x ) = x 2 - x :‬‬ ‫ﻓﻨﺠﺩ ‪:‬‬ ‫‪g‬‬ ‫(‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫‪4‬‬ ‫;‬ ‫‪h(x)=x+3‬‬ ‫‪x‬‬ ‫) ‪( U1‬‬ ‫) ‪(U3) (U2‬‬‫* ﺤل ﺒﻴﺎﻨﻴﺎ ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f ( x ) = - 2 :‬؛ ∆ ﻤﺴﺘﻘﻴﻡ ﻤﻌﺎﺩﻟﺘﻪ ‪( )y = −2 :‬‬ ‫ﻨﻼﺤﻅ ﺒﻴﺎﻨﻴﺎ ﺃﻥ ‪ U1 ∩ ∆ = φ :‬ﻭﻋﻠﻴﻪ ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ‪( ) ( ).‬‬ ‫* ﺤل ﺒﻴﺎﻨﻴﺎ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪f ( x ) > - 2 :‬‬

‫ﺒﻴﺎﻥ ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﻤﻨﺤﻨﻰ )‪ (U1‬ﻴﻘﻊ ﻓﻭﻕ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل) (‬ ‫ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪. R‬‬ ‫‪4‬‬ ‫≤‬ ‫‪x‬‬ ‫‪+3‬‬ ‫‪:‬‬ ‫ﻟﻠﻤﺘﺭﺍﺠﺤﺔ‬ ‫ﺍﻟﺤﺴﺎﺒﻲ‬ ‫ﺍﻟﺤل‬ ‫*‬ ‫‪x‬‬ ‫‪4‬‬ ‫‪-‬‬ ‫‪x2‬‬ ‫‪-‬‬ ‫‪3x‬‬ ‫≤‬ ‫‪0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪4‬‬ ‫‪-‬‬ ‫(‬ ‫‪x‬‬ ‫‪+‬‬ ‫)‪3‬‬ ‫≤‬ ‫‪0‬‬ ‫‪:‬‬ ‫ﺃﻥ‬ ‫ﻴﻌﻨﻲ‬ ‫ﻭﻫﺫﺍ‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪x2‬‬ ‫‪- 3x‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫≤‬ ‫‪0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪x‬‬ ‫ـ ﻨﺤﻠل ﺍﻟﻌﺒﺎﺭﺓ ‪− x2 - 3x + 4 :‬‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = b2 - 4ac = 9 - 4( - 1 ) ( 4 ) = 25 :‬‬ ‫‪x1‬‬ ‫=‬ ‫‪3-5‬‬ ‫=‬ ‫‪1‬‬ ‫;‬ ‫‪x2‬‬ ‫=‬ ‫‪3+5‬‬ ‫=‬ ‫‪-‬‬ ‫‪4‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪-2‬‬ ‫‪-2‬‬ ‫ﻤﻼﺤﻅﺔ ﻫﺎﻤﺔ ‪\":‬‬‫ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ a x2 + b x + c = 0‬ﻓﻲ ﺤﺎﻟﺔ ‪a + c = - b‬‬ ‫‪c‬‬ ‫ﺘﻘﺒل ﺤﻠﻴﻥ ﻫﻤﺎ ‪ 1 :‬ﻭ ‪. a‬‬ ‫ﺇﺫﻥ ‪- x 2 - 3x + 4 = - ( x -1) ( x + 4 ) :‬‬ ‫‪−‬‬ ‫‪(x‬‬ ‫‪−‬‬ ‫‪1)(x‬‬ ‫‪+‬‬ ‫)‪4‬‬ ‫≤‬ ‫‪0‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺘﺼﺒﺢ ‪:‬‬ ‫‪x‬‬ ‫ﺠﺩﻭل ﺍﻹﺸﺎﺭﺓ ‪:‬‬ ‫‪x‬‬ ‫∞‪-∞ -4 0 1 +‬‬ ‫)‪-(x–1‬‬ ‫‪+++ -‬‬ ‫‪x+4‬‬ ‫‪x‬‬ ‫‪- +++‬‬ ‫)‪-(x–1)(x+4‬‬ ‫‪- - ++‬‬‫) ‪− ( x -1) ( x + 4‬‬ ‫‪- ++ -‬‬ ‫‪+-+-‬‬‫‪x‬‬ ‫ﻭﻋﻠﻴﻪ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪[ [ [ [E1 = - 4 , 0 U 1, + ∞ :‬‬

‫ﺃﻱ ‪g ( x ) ≤ h ( x ) :‬‬ ‫‪4‬‬ ‫≤‬ ‫‪x‬‬ ‫‪+3‬‬ ‫‪:‬‬ ‫ﻟﻠﻤﺘﺭﺍﺠﺤﺔ‬ ‫ﺍﻟﺒﻴﺎﻨﻲ‬ ‫ﺍﻟﺤل‬ ‫*‬ ‫‪x‬‬ ‫ﻭﻫﺫﺍ ﻴﺅﻭل ﺇﻟﻰ ﺇﻴﺠﺎﺩ ﻭﻀﻌﻴﺔ ﺍﻟﻤﻨﺤﻨﻲ ‪ U2‬ﻭ ‪( ) ( ). U3‬‬‫ﻤﻥ ﺨﻼل ﺍﻟﺒﻴﺎﻨﻲ ﻨﻼﺤﻅ ﺃﻥ ‪ U2 :‬ﻴﻘﻊ ﺘﺤﺕ ‪ U3‬ﻓﻲ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴـﻥ ‪( ) ( ):‬‬ ‫[ ‪ ] - 4 , 0‬ﻭ [ ∞ ‪ ]1, +‬ﻭﻋﻠﻴﻪ ) ‪. g ( x ) < h ( x‬‬ ‫ﻜﺫﻟﻙ ‪ U2‬ﻴﻘﻁﻊ ‪ U3‬ﻓﻲ ﺍﻟﻨﻘﻁﺘﻴﻥ ﺍﻟﻠﺘﻴﻥ ﻓﺎﺼﻠﺘﻬﻤﺎ ‪ 1، - 4‬ﻭﻋﻠﻲ ﻤﺠﻤﻭﻋﺔ) ( ) (‬‫ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ) ‪ g ( x ) ≤ f ( x‬ﻫﻲ ‪E1 = [ - 4 , 0 [U [ 1, + ∞ [ :‬‬ ‫* ﺍﻟﺤل ﺍﻟﺤﺴﺎﺒﻲ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪x + 3 ≥ x 2 - x :‬‬‫ﻭﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ‪ x + 3 - x2 + x ≥ 0 :‬ﺃﻱ ‪- x2 + 2 x + 3 ≥ 0 :‬‬ ‫ﻟﺩﻴﻨﺎ ‪∆ = 4 + 12 = 16 :‬‬ ‫‪x1‬‬ ‫=‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪4‬‬ ‫=‬ ‫‪3‬‬ ‫;‬ ‫‪x2‬‬ ‫=‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫=‬ ‫‪-1‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪2‬‬ ‫ﻤﻼﺤﻅﺔ ﻫﺎﻤﺔ ‪\":‬‬‫‪ a x2 + b x + c = 0‬ﻓﻲ ﺤﺎﻟﺔ ‪a + c = b‬‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫‪−c‬‬ ‫ﺘﻘﺒل ﺤﻠﻴﻥ ﻫﻤﺎ ‪ - 1 :‬ﻭ ‪. a‬‬ ‫ﻭﻋﻠﻴﻪ ‪- x 2 + 2x + 3 = - ( x - 3 ) ( x + 1 ) :‬‬ ‫ﻭ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺘﺼﺒﺢ ‪- ( x - 3 ) ( x + 1 ) ≥ 0 :‬‬ ‫ﺠﺩﻭل ﺍﻹﺸﺎﺭﺓ ‪:‬‬ ‫‪x‬‬ ‫∞‪−‬‬ ‫‪-1‬‬ ‫∞‪3 +‬‬ ‫)‪-(x–3‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪x+1‬‬ ‫‪-‬‬ ‫‪+‬‬‫)‪-(x–3)(x+1‬‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪-‬‬ ‫‪-‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪[ ]E2 = -1, 3 :‬‬‫* ﺍﻟﺤل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﻤﺘﺭﺍﺠﺤﺔ ‪ x + 3 ≥ x 2 - x :‬ﺃﻱ ‪h ( x ) ≥ f ( x ) :‬‬ ‫ﻭﻫﺫﺍ ﻴﺅﻭل ﺇﻟﻰ ﺩﺭﺍﺴﺔ ﺍﻟﻭﻀﻌﻴﺔ ﻟﻜل ﻤﻥ ‪ U1‬ﻭ ‪( ) ( ). U3‬‬

‫ﻤﻥ ﺨﻼل ﺍﻟﺒﻴﺎﻨﻲ ﻨﻼﺤﻅ ﺃﻥ ‪ U3‬ﻴﻘﻁﻊ ﻓﻭﻕ ‪ U1‬ﻓﻲ ﺍﻟﻤﺠﺎل ‪] [ ( ) ( )-1, 3‬‬ ‫ﻭﻴﻘﻁﻌﻪ ﻋﻨﺩ ﺍﻟﻨﻘﻁﺘﻴﻥ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺘﻴﻥ ‪– 1 ، 3‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ ‪[ ]E 2 = -1, 3 :‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 20‬‬‫‪. × (4‬‬ ‫‪. × (3‬‬ ‫‪. × (2‬‬ ‫‪. 9 (1‬‬‫‪(7 . 9 (6‬‬ ‫‪. 9 (5‬‬ ‫‪. 9 (8‬‬ ‫×‪.‬‬ ‫ﺘﻤﺭﻴﻥ ‪: 21‬‬ ‫ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ‪: f‬‬ ‫‪-π‬‬ ‫‪π‬‬‫‪−π 2‬‬ ‫‪π‬‬ ‫‪2‬‬ ‫ﻨﺭﺴﻡ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﺘﻲ ﻤﻌﺎﺩﻻﺘﻬﺎ ‪ y = -1 , y = 1 , y = 0 :‬ﻓﻨﺠﺩ ‪.‬‬ ‫‪ (1‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f (x) = 0‬ﺜﻼﺜﺔ ﺤﻠﻭل ﻫﻲ ‪− π , 0 , π :‬‬ ‫‪π‬‬ ‫‪ (2‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f (x) = 1‬ﺤل ﻤﻀﺎﻋﻑ ﻫﻭ ‪2 :‬‬ ‫‪−π‬‬ ‫‪ (3‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ f (x) = -1‬ﺤل ﻤﻀﺎﻋﻑ ﻫﻭ ‪2 :‬‬ ‫‪ -‬ﺍﺴﺘﻨﺘﺎﺝ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ‪:‬‬‫‪ (1‬ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) ≥ 0‬ﻫﻲ ﻗﻴﻡ ‪ x‬ﺍﻟﺘﻲ ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﺎل ‪[ ]. 0 , π‬‬ ‫‪ (2‬ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ f (x) f 1‬ﻟﻴﺱ ﻟﻬﺎ ﺤﻠﻭل ‪.‬‬