دروس مادة الرياضيات للفصل الاول سنة رابعة متوسط

‫ﺍﻟﺘﻤﺎﺭﻴﻥ‬ ‫‪ MNP /1‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪. P‬‬ ‫ﺍﻨﻘل ﻭ ﺍﺘﻤﻡ ﺍﻟﻤﺴﺎﻭﻴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪PM‬‬ ‫*‪= cos…/‬‬ ‫‪PN‬‬ ‫*‪= cos… = sin…/‬‬ ‫‪MN‬‬ ‫‪MN‬‬ ‫‪PM‬‬ ‫‪= ….‬‬ ‫*‪/‬‬ ‫‪PN‬‬ ‫*‪= …. /‬‬ ‫‪PN‬‬ ‫‪PM‬‬ ‫‪ RST /2‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪.R‬‬ ‫ﺍﻨﻘل ﻭ ﺍﺘﻤﻡ ﺍﻟﺠﻤل ﺍﻟﺘﺎﻟﻴﺔ ﺏ ‪.Tan , Sin , Cos‬‬ ‫*‪/‬ﺍﺫﺍ ﻋﻠﻤﻨﺎ ﻁﻭل ‪ RT , RS‬ﻨﺴﺘﻁﻴﻊ ﺤﺴﺎ‪ ...‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪S‬‬ ‫ﻭﺍﻟﺯﺍﻭﻴﺔ ˆ‪T‬‬ ‫*‪/‬ﻟﻤﻌﺭﻓﺔ ﻁﻭل‪ RS‬ﻭ ‪ RT‬ﻨﺤﺴﺏ ‪ ...‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪ S‬ﻭ ﺍﻟﺯﺍﻭﻴﺔ ˆ‪. T‬‬ ‫*‪ /‬ﺍﺫﺍ ﻋﻠﻤﻨﺎ ﺍﻟﻁﻭﻟﻴﻥ ‪ ST‬ﻭ‪ RT‬ﻨﺤﺴﺏ ‪ ...‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪ S‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪. T‬‬ ‫ﻟـ‪:‬‬ ‫‪1‬‬ ‫ﺍﻟﻤﺩﻭﺭ‬ ‫ﺃﻭ‬ ‫‪/3‬ﺒﺎﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﺃﻋﻁ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺘﺎﻤﺔ‬ ‫‪1000‬‬ ‫‪Tan 45° , Sin 40° , Cos 35°‬‬ ‫‪Tan 90° , Cos 50° , Sin 65°‬‬‫‪ /4‬ﺒﺎﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﺃﻋﻁ ﻓﻲ ﻜل ﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﺩﻭﺭ ﺇﻟﻰ ﺍﻟﻭﺤﺩﺓ ﺒﺎﻟﺩﺭﺠﺎﺕ ﻟﻘﻴﺎﺴﺎﺕ ﺍﻟﺯﺍﻭﻴﺔ ˆ‪. A‬‬‫‪Tan‬‬ ‫= ˆ‪A‬‬ ‫‪5‬‬ ‫*‪/‬‬ ‫‪Sin‬‬ ‫= ˆ‪A‬‬ ‫‪0.9‬‬ ‫*‪/‬‬ ‫‪Cos‬‬ ‫ˆ‪A‬‬ ‫=‬ ‫‪1‬‬ ‫*‪/‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪ MNP/5‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪ M‬ﺤﻴﺙ‪ Nˆ = 72°:‬ﻭ ‪MP = 6.3 cm‬‬ ‫‪1‬‬ ‫ﺜﻡ ﺃﻋﻁ ﻤﺩﻭﺭﻩ ﺇﻟﻰ‬ ‫‪NP‬‬ ‫ﺃﺤﺴﺏ‬ ‫‪. 100‬‬ ‫‪ ABC/6‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪ A‬ﺤﻴﺙ ‪ Bˆ = 68° :‬ﻭ ‪BC = 5.3 cm‬‬ ‫‪.‬‬ ‫‪1‬‬ ‫ﺇﻟﻰ‬ ‫ﻤﺩﻭﺭﻩ‬ ‫ﺃﻋﻁ‬ ‫ﺜﻡ‬ ‫‪AC‬‬ ‫ﺃﺤﺴﺏ‬ ‫‪100‬‬

‫‪ UNE /7‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ‪ U‬ﻓﻲ ﺤﻴﺙ‪ Nˆ = 20°:‬ﻭ ‪UN = 4.5cm‬‬ ‫‪.‬‬ ‫‪1‬‬ ‫ﺇﻟﻰ‬ ‫ﻤﺩﻭﺭﻩ‬ ‫ﺃﻋﻁ‬ ‫ﺜﻡ‬ ‫‪EU‬‬ ‫ﺃﺤﺴﺏ‬ ‫‪100‬‬‫‪ ABC /8‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ‪ A‬ﻓﻲ ﺤﻴﺙ ‪BC= 11.5cm , AC = 7.8 cm :‬‬ ‫ˆ‪B‬‬ ‫ﻟﻘﻴﺱ ﺍﻟﺯﺍﻭﻴﺔ‬ ‫‪1‬‬ ‫ﺃﻋﻁ ﺍﻟﻤﺩﻭﺭ ﺇﻟﻰ ‪10‬‬‫‪ RST /9‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ‪ T‬ﻓﻲ ﺤﻴﺙ‪RS = 7.5 cm , RT = 4.7 cm :‬‬‫ﺍﻟﻭﺤﺩﺓ‪.‬‬ ‫‪1‬‬ ‫ﻤﺩﻭﺭﻩ‬ ‫ﺃﻋﻁ‬ ‫ﺜﻡ‬ ‫ﺃﺤﺴﺏ ˆ‪R‬‬ ‫ﺇﻟﻰ ‪100‬‬‫‪ EFG /10‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ‪ E‬ﻓﻲ ﺤﻴﺙ‪EF = 5 cm , EG = 3.5 cm :‬‬ ‫‪.‬‬ ‫‪1‬‬ ‫ˆ‪ G‬ﺇﻟﻰ‬ ‫ﺃﻋﻁ ﻤﺩﻭﺭ ﻗﻴﺱ ﺍﻟﺯﺍﻭﻴﺔ‬ ‫‪10‬‬‫‪ ABC /11‬ﻤﺜﻠﺙ ‪ ،‬ﺍﻻﺭﺘﻔﺎﻉ ]‪ [HA‬ﺍﻟﻤﺘﻌﻠﻕ ﺒﺎﻟﻀﻠﻊ ]‪ [BC‬ﺤﻴﺙ‪:‬‬‫‪AH = 4.9cm , BH = 1.4cm , CH = 2cm‬‬ ‫‪1‬‬‫ﺃﺤﺴﺏ ﺍﻟﻤﺩﻭﺭ ﺇﻟﻰ ‪ 10‬ﻟﻘﻴﺱ ﺯﺍﻭﻴﺔ ﻤﻥ ﺯﻭﺍﻴﺎ ﺍﻟﻤﺜﻠﺙ ‪.‬‬‫ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻌﻼﻗﺔ ‪Cos2 x + Sun2 x= 1 :‬‬ ‫= ‪Cos‬‬ ‫‪3‬‬ ‫‪ /12‬ﻋﻠﻤﺎ ﺃﻥ ‪:‬‬ ‫‪5‬‬ ‫‪ -‬ﺃﺤﺴﺏ ‪ Sin x‬ﻭ ‪.Tan x‬‬ ‫= ‪Sin‬‬ ‫‪8‬‬ ‫ﺃﻥ‪:‬‬ ‫ﻋﻠﻤﺎ‬ ‫‪/13‬‬ ‫‪17‬‬ ‫‪ -‬ﺃﺤﺴﺏ ‪ Cos x‬ﻭ ‪.Tan x‬‬‫‪Tan‬‬ ‫‪x‬‬ ‫=‬ ‫‪5‬‬ ‫ﻭ‬ ‫= ‪Sin x‬‬ ‫‪5‬‬ ‫‪ /14‬ﻋﻠﻤﺎ ﺃﻥ ‪:‬‬ ‫‪12‬‬ ‫‪13‬‬ ‫‪ -‬ﺃﺤﺴﺏ ‪.Cos x‬‬‫‪ -‬ﺘﺤﻘﻕ ﻤﻥ ﺼﺤﺔ ﺍﻟﻌﻼﻗﺔ ‪Cos2 x + Sin2 x = 1 :‬‬

‫‪ AB = 10‬ﻭ = ‪BH‬‬ ‫‪ ABC /15‬ﻤﺜﻠﺙ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ ﺭﺃﺴﻪ ﺍﻷﺴﺎﺴﻲ ‪.A‬‬ ‫*‪ [BH]-‬ﻭ ]‪ [CK‬ﺍﻻﺭﺘﻔﺎﻋﻴﻥ ﺍﻟﻤﺘﻌﻠﻘﻴﻥ ﺒﺎﻟﻀﻠﻌﻴﻥ]‪ [AC‬ﻭ ]‪ ،[AB‬ﺤﻴﺙ‬ ‫‪ 6‬ﻭ ‪.AK = 8‬‬ ‫*‪ -‬ﺃﻋﻁ ﺍﻟﻘﻴﻤﺔ ﻋﻠﻰ ﺸﻜل ﻜﺴﺭ ﻏﻴﺭ ﻗﺎﺒل ﻟﻼﺨﺘﺯﺍل ﻟـ‪:‬‬ ‫ˆ‪. Cos Aˆ , Sin Aˆ ,Tan A‬‬

‫ﺍﻟﻤﺴﺎﺌل‬ ‫‪ /16‬ﺍﻨﻜﺴﺎﺭ ﺍﻟﻀﻭﺀ‪:‬‬‫ﻴﻨﺤﺭﻑ ﺸﻌﺎﻉ ﻀﻭﺌﻲ ﻋﻨﺩﻤﺎ ﻴﻨﺘﻘل ﻤﻥ ﺍﻟﻬﻭﺍﺀ ﺇﻟﻰ ﺍﻟﻤﺎﺀ‪.‬‬‫ﺍﻟﺯﺍﻭﻴﺔ ˆ‪ i‬ﺘﺴﻤﻰ )ﺯﺍﻭﻴﺔ ﺍﻟﻭﺭﻭﺩ( ﻭﺍﻟﺯﺍﻭﻴﺔ‬‫ˆ‪ r‬ﺘﺴﻤﻰ )ﺯﺍﻭﻴﺔ ﺍﻻﻨﻜﺴﺎﺭ( ﺍﻟﻤﻭﻀﺤﺔ ﻓﻲ ﺍﻟﺸﻜل‬‫= ˆ‪Sin r‬‬ ‫‪3‬‬ ‫‪Sin‬‬ ‫ﺤﻴﺙ‪iˆ :‬‬ ‫‪4‬‬‫ﺃ ( ﺃﺤﺴﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﺍﻟﻰ ﺍﻟﻭﺤﺩﺓ‬ ‫ﻟـ ˆ‪ r‬ﻋﻠﻤﺎ ﺃﻥ ‪. iˆ =35°‬‬‫ﺏ( ﺃﺤﺴﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﺍﻟﻰ ﺍﻟﻭﺤﺩﺓ‬ ‫ﻟـ ˆ‪ i‬ﻋﻠﻤﺎ ﺃﻥ ‪. rˆ =45°‬‬ ‫‪ /17‬ﺇﺭﺘﻔﺎﻉ ﻋﻤـﻭﺩ‬‫ﻟﺤﺴﺎﺏ ﺍﺭﺘﻔﺎﻉ ‪ h‬ﻟﻌﻤﻭﺩ ﻗﻤﻨﺎ ﺒﺎﻟﻘﻴﺎﺴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪a = 58.5 , b = 35.1‬‬ ‫ﻭ ‪AB = 18.7m‬‬‫ﺃ ( ﻓﻲ ﺍﻟﻤﺜﻠﺙ ‪ AHS‬ﻭ ‪BHS‬‬‫ﻋﺒﺭﻋﻥ ‪ AH‬ﻭ ‪ BH‬ﺒﺩﻻﻟﺔ ‪. h‬‬‫ﺏ( ﺍﺴﺘﻨﺘﺞ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﺍﻟﻰ ﺍﻟﻭﺤﺩﺓ ﻟـ ‪.h‬‬ ‫‪ /18‬ﻁﻭل ﻭﺘﺭ ﻓﻲ ﺩﺍﺌﺭﺓ‪.‬‬‫ﺍﻟﻨﻘﻁ ‪ A‬ﻭ ‪b‬ﻴﻨﺘﻤﻴﺎﻥ ﺍﻟﻰ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ‪ θ‬ﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪8cm‬‬ ‫ﻭ‪AθˆB =40°‬‬ ‫ﺃ ( ﺃﻨﺸﺊ ﺍﻟﺸﻜل ﻭﺍﺭﺴﻡ ﺍﻻﺭﺘﻔﺎﻉ‬ ‫]‪ [OH‬ﺍﻟﻤﺘﻌﻠﻕ ﺒﺎﻟﻀﻠﻊ ]‪.[AB‬‬ ‫ﺏ( ﺃﺤﺴﺏ ﻗﻴﻤﺔ ‪، AB‬‬ ‫‪1‬‬ ‫ﺜﻡ ﺃﻋﻁ ﻗﻴﻤﺘﻬﺎ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﺇﻟﻰ ‪. 10‬‬

‫ﺤل ﺍﻟﺘﻤﺎﺭﻴﻥ ﻭ ﺍﻟﻤﺴﺎﺌل‬ ‫ﺇﺘﻤﺎﻡ ﺍﻟﻤﺴﺎﻭﻴﺎﺕ‪:‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل‪:‬‬ ‫*‪/‬‬‫‪PN‬‬ ‫‪= Cos‬‬ ‫ˆ‪N‬‬ ‫ˆ‪= sin M‬‬‫‪MN‬‬ ‫‪PM‬‬ ‫‪= Cos‬‬ ‫ˆ‪M‬‬ ‫*‪/‬‬ ‫‪MN‬‬ ‫‪PM‬‬ ‫‪= Tang‬‬ ‫ˆ‪N‬‬ ‫*‪/‬‬ ‫‪PN‬‬ ‫‪PN‬‬ ‫‪= Tang‬‬ ‫ˆ‪M‬‬ ‫*‪/‬‬ ‫‪PM‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ‪ RST :‬ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪.R‬‬ ‫ﺇﺘﻤﺎﻡ ﺍﻟﺠﻤل ﺒـ ‪Tang , Sin , Cos :‬‬ ‫*‪ /‬ﺇﺫﺍ ﻋﻠﻤﻨﺎ ﻁﻭل ‪ RS‬ﻭ ‪ RT‬ﻨﺤﺴﺏ‬ ‫‪Tang‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪ S‬ﻭ ‪ Tang‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪.T‬‬ ‫*‪ /‬ﺒﻤﻌﺭﻓﺔ ﻁﻭل‪ RS‬ﻭ ‪ ST‬ﻨﺤﺴﺏ‬ ‫‪ Cos‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪ S‬ﻭ ‪ Sin‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪. T‬‬‫*‪ /‬ﺇﺫﺍ ﻋﻠﻤﻨﺎ ﺍﻟﻁﻭل ‪ ST‬ﻭ ‪ RT‬ﻨﺤﺴﺏ ‪ Sin‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪ S‬ﻭ‪ Cos‬ﺍﻟﺯﺍﻭﻴﺔ ˆ‪. T‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻟﺙ‪ :‬ﺒﺎﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﻨﺠﺩ‪:‬‬‫‪Tang 45°= 1, Sin 40°= 0.6428, Cos 35°= 0.8192‬‬‫‪Tang 90°= , Cos 50°= 0.6428, Sin 65°= 0.9063‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺭﺍﺒﻊ‪ :‬ﺒﺎﺴﺘﻌﻤﺎل ﺍﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﺔ ﻨﺠﺩ‪:‬‬ ‫‪Aˆ =71°‬‬ ‫ﻤﻌﻨﺎﻩ‪:‬‬ ‫‪Cos‬‬ ‫ˆ‪A‬‬ ‫‪1‬‬ ‫‪=3‬‬ ‫‪ Sin Aˆ = 0.9‬ﻤﻌﻨﺎﻩ ‪Aˆ =71° :‬‬ ‫ﻤﻌﻨﺎﻩ‪Aˆ = 59° :‬‬ ‫‪Tang‬‬ ‫ˆ‪A‬‬ ‫=‬ ‫‪5‬‬ ‫‪3‬‬

‫‪1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺨﺎﻤﺱ‪ :‬ﺤﺴﺎﺏ ‪ NP‬ﺒﺎﻟﺘﺩﻭﻴﺭ ﺇﻟﻰ ‪: 100‬‬‫= ‪NP‬‬ ‫‪6.3 MP‬‬ ‫ﺃﻱ‬ ‫ˆ‪Sin N‬‬ ‫=‬ ‫‪MP‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫ˆ‪ NP = SinN‬ﻭ ﻤﻨﻪ ‪Sin72°‬‬ ‫‪NP‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪.NP = 6.62 cm :‬‬ ‫‪1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺴﺎﺩﺱ‪ :‬ﺤﺴﺎﺏ ‪ AC‬ﺒﺎﻟﺘﺩﻭﻴﺭ ﺇﻟﻰ ‪: 100‬‬ ‫‪ AC = BC x Sin‬ﻭ ﻤﻨﻪ‬ ‫ﺃﻱ ˆ‪B‬‬ ‫ˆ‪Sin B‬‬ ‫=‬ ‫‪AC‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪BC‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪.AC = 4.91 cm :‬‬ ‫‪AC = 5.3 x Sin 68‬‬ ‫‪:‬‬ ‫‪1‬‬ ‫ﺒﺎﻟﺘﺩﻭﻴﺭ ﺇﻟﻰ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺴﺎﺒﻊ‪ :‬ﺤﺴﺎﺏ ‪EU‬‬ ‫‪100‬‬‫‪ EU = UN x Tang‬ﻭ ﻤﻨﻪ‬ ‫ﺃﻱ ˆ‪N‬‬ ‫ˆ‪Tang N‬‬ ‫=‬ ‫‪EU‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪UN‬‬ ‫‪ EU = 5 x Tang 20°‬ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪EU = 1.82 cm :‬‬ ‫‪:‬‬ ‫ˆ‪B‬‬ ‫ﻟﻘﻴﺱ ﺍﻟﺯﺍﻭﻴﺔ‬ ‫‪1‬‬ ‫ﺇﻋﻁﺎﺀ ﺍﻟﻤﺩﻭﺭ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻤﻥ‪:‬‬ ‫ﺇﻟﻰ ‪10‬‬‫ﻭﻤﻨﻪ ‪Bˆ = 42.5°‬‬ ‫= ˆ‪Sin B‬‬ ‫‪4.7‬‬ ‫ﺃﻱ‬ ‫= ˆ‪Sin B‬‬ ‫‪AC‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪7.5‬‬ ‫‪BC‬‬ ‫‪:‬‬ ‫ˆ‪R‬‬ ‫ﻟﻘﻴﺱ ﺍﻟﺯﺍﻭﻴﺔ‬ ‫‪1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺘﺎﺴﻊ‪:‬‬ ‫ﺇﻋﻁﺎﺀ ﺍﻟﻤﺩﻭﺭ ﺇﻟﻰ ‪100‬‬‫ﻭﻤﻨﻪ ‪Rˆ = 42.5°‬‬ ‫= ˆ‪Sin R‬‬ ‫‪4.7‬‬ ‫ﺃﻱ‬ ‫ˆ‪Sin R‬‬ ‫=‬ ‫‪TR‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪7.5‬‬ ‫‪RS‬‬ ‫‪:‬‬ ‫ˆ‪G‬‬ ‫ﻟﻘﻴﺱ ﺍﻟﺯﺍﻭﻴﺔ‬ ‫‪1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﻌﺎﺸﺭ‪:‬‬ ‫ﺇﻋﻁﺎﺀ ﺍﻟﻤﺩﻭﺭ ﺇﻟﻰ ‪10‬‬‫ﻭﻤﻨﻪ ‪Gˆ = 55°,..‬‬ ‫ˆ‪Sin G‬‬ ‫=‬ ‫‪5‬‬ ‫ﺃﻱ‬ ‫ˆ‪Sin G‬‬ ‫=‬ ‫‪EF‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪3.5‬‬ ‫‪EG‬‬

: ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﺤﺩ ﻋﺸﺭ‬ Â = 180°-……- …… ; : ‫ﻟﺩﻴﻨﺎ‬ Â = 180°- Bˆ - Gˆ A ≈ : ‫ﻭﻤﻨﻪ‬ : 12 ‫ﺍﻟﺘﻤﺭﻴﻥ‬ Cos2 x + Sin2 x = 1 ‫ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻌﻼﻗﺔ‬ : ‫ﻟﺩﻴﻨﺎ‬ Sin2 x = 1 - Cos2 x sin ²x = 1−  5 ² = 1 − 95 : ‫ﺃﻱ‬  5  25 sin ²x 16 25 sin = 4 : ‫ﺃﻱ‬ 5 4Tang = 4 < Tangx = 5 , Tangx = sin x ‫ﻟﺩﻴﻨﺎ‬ 3 3 cos x 5

‫اﻟﺤﺴﺎب اﻟﺤﺮﻓﻲ‬ ‫اﻟﻜﻔﺎءات اﻟﻤﺴﺘﻬﺪﻓﺔ‪:‬‬ ‫*‪/‬ﻤﻌﺭﻓﺔ ﺍﻟﻤﺘﻁﺎﺒﻘﺎﺕ ﺍﻟﺸﻬﻴﺭﺓ ﻭﺘﻭﻅﻴﻔﻬﺎ ﻓﻲ ﺍﻟﺤﺴﺎﺏ‬ ‫ﺍﻟﻤﺘﻤﻌﻥ ﻓﻴﻪ ﻭ ﻓﻲ ﺍﻟﻨﺸﺭ ﻭﺍﻟﺘﺤﻠﻴل‪.‬‬ ‫*‪/‬ﻨﺸﺭ ﺃﻭ ﺘﺤﻠﻴل ﻋﺒﺎﺭﺍﺕ ﺠﺒﺭﻴﺔ ﺒﺴﻴﻁﺔ‪.‬‬ ‫*‪ /‬ﺤل ﻤﻌﺎﺩﻻﺕ ﻴﺅﻭل ﺤﻠﻬﺎ ﺇﻟﻰ ﺤل \"ﻤﻌﺎﺩﻟﺔ ﺠﺩﺍﺀ\"‪.‬‬ ‫*‪/‬ﺤل ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭﻟﻴﻥ‬ ‫ﺠﺒﺭﻴﺎ‪.‬‬ ‫*‪ /‬ﺘﻔﺴﻴﺭ ﺤل ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ‬ ‫ﺒﻤﺠﻬﻭﻟﻴﻥ ﺒﻴﺎﻨﻴﺎ‪.‬‬ ‫*‪ /‬ﺤل ﻤﺘﺭﺍﺠﺤﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭل ﻭﺍﺤﺩ ﻭ ﺘﻤﺜﻴل ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭﻟﻬﺎ ﻋﻠﻰ‬ ‫ﻤﺴﺘﻘﻴﻡ ﻤﺩﺭﺝ‪.‬‬‫*‪ /‬ﺤل ﻤﺸﻜﻼﺕ ﺒﺘﻭﻅﻴﻑ ﻤﻌﺎﺩﻻﺕ ﺃﻭ ﻤﺘﺭﺍﺠﺤﺎﺕ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭل ﻭﺍﺤﺩ ﺃﻭ‬ ‫ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭﻟﻴﻥ‪.‬‬‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫ﺍﻟﺩﺭﺱ‬

‫ﺍﻟﺩﺭﺱ‬ ‫‪ .1‬ﺍﻟﻨـﺸــﺭ‪:‬‬ ‫ﺘﻌﺭﻴﻑ‪:‬‬ ‫ﻨﺸﺭ ﺠﺩﺍﺀ ‪ :‬ﻫﻭ ﺘﺤﻭﻴﻠﻪ ﺇﻟﻰ ﻤﺠﻤﻭﻉ ﺃﻭ ﺇﻟﻰ ﻓﺭﻕ‪.‬‬ ‫ﺨﻭﺍﺹ‪:‬‬ ‫‪ ,a , b, c, d , k‬ﺃﻋﺩﺍﺩ‬ ‫‪k (a + b) = ka +kb.‬‬‫‪K ( a – b) = ka – kb.‬‬‫‪( a + b )( c + d ) = ac + ad + bc + bd .‬‬ ‫ﺃﻤﺜﻠﺔ‪ :‬ﺃﻨﺸﺭ ﻭ ﺒﺴﻁ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫‪A = 3 ( 2 + 5X ) =3 × 2 + 3 × 5X = 6 + 5X.‬‬‫)‪B = ( X + 4 )( X + 3) = (X × X)+(X× 3) + (4 × X) + (4 × 3‬‬‫‪= X2 + 3X + 4X +12‬‬ ‫ﺘﺒﺴﻴﻁ ﺍﻟﻌﺒﺎﺭﺓ ‪3X+4X :‬‬‫‪B = X2 + 7X + 12.‬‬

‫‪( a + b )2 = a2 + 2 ab + b2‬‬ ‫‪ .2‬ﺍﻟﻤﺘﻁﺎﺒﻘﺎﺕ ﺍﻟﺸﻬﻴﺭﺓ‪:‬‬ ‫ﺍﻟﺤﺩ ‪ 2 ab‬ﻫﻭ ﻀﻌﻑ ﺍﻟﺠﺩﺍﺀ ﻟـ ‪ a‬ﻭ ‪b‬‬ ‫*ﻤﺭﺒﻊ ﻤﺠﻤﻭﻉ ‪:‬‬‫‪A = ( x + 3)2‬‬ ‫ﻤﺜﺎل‪:‬‬ ‫ﺃﻨﺸﺭ ﺍﻟﻌﺒﺎﺭﺓ ‪ A‬ﺤﻴﺙ ‪:‬‬ ‫‪= x2 + 2(x × 3) + 32‬‬ ‫* ﻤﺭﺒﻊ ﻓﺭﻕ‪:‬‬‫‪A = x2 + 6x + 9.‬‬ ‫ﻤﺜﺎل‪:‬‬ ‫‪( a + b )2 = a2 + 2ab +b2‬‬ ‫ﺃﻨﺸﺭ ﺍﻟﻌﺒﺎﺭﺓ ‪ A‬ﺤﻴﺙ ‪:‬‬ ‫‪A = ( x – 4 )2‬‬ ‫*ﺠﺩﺍﺀ ﻤﺠﻤﻭﻉ ﻭﻓﺭﻕ‪:‬‬ ‫‪= x2 – 2 (4 × x) + 42‬‬ ‫‪A = x2 – 8x + 16.‬‬ ‫‪( a + b)( a – b ) = a2 – b2.‬‬ ‫ﻤﺜﺎل‪ :‬ﺃﻨﺸﺭ ﺍﻟﻌﺒﺎﺭﺓ ‪ A‬ﺤﻴﺙ ‪:‬‬‫)‪A = ( x + 2 )( x – 2‬‬ ‫‪= x2 – (2)2‬‬‫‪A = x2 – 4.‬‬

‫‪ .3‬ﺍﻟﺘﺤﻠﻴل ‪:‬‬ ‫ﺘﻌﺭﻴﻑ‪:‬‬ ‫ﺘﺤﻠﻴل ﻤﺠﻤﻭﻉ ﺃﻭ ﻓﺭﻕ ﻫﻭ ﺘﺤﻭﻴﻠﻪ ﺇﻟﻰ ﺠﺩﺍﺀ ﺃﻭ ﻓﺭﻕ‪.‬‬ ‫ﻜﻲ ﻨﺘﻤﻜﻥ ﻤﻥ ﺘﺤﻠﻴل ‪ ،‬ﻨﺒﺤﺙ ﻋﻥ ﺍﻟﻌﺎﻤل ﺍﻟﻤﺸﺘﺭﻙ‪،‬‬ ‫ﻓﺈﻥ ﻟﻡ ﻴﻅﻬﺭ ﺍﻟﻌﺎﻤل ﺍﻟﻤﺸﺘﺭﻙ ﺠﺭﺏ ﺇﺤﺩﻯ ﺍﻟﻤﺘﻁﺎﺒﻘﺎﺕ ﺍﻟﺸﻬﻴﺭﺓ‪.‬‬ ‫ﺨـﻭﺍﺹ‪:‬‬ ‫‪Ka + kb = k ( a + b ).‬‬‫‪Ka – kb = k ( a – b ).‬‬‫‪a2 + 2ab +b2 = ( a + b )2‬‬‫‪a2 - 2ab +b2 = ( a - b )2 .‬‬‫‪a2 - b2 = ( a + b ) ( a – b ).‬‬ ‫ﺃﻤﺜﻠﺔ‪ :‬ﺤﻠل ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫‪A = 4x + 12 =4x + 4× 3‬‬‫‪A = 4 ( x + 3 ).‬‬‫‪B = (x + 2)(2x +1) – (x +2 ) x‬‬‫)‪= ( x + 2 )( 2x +1 – x‬‬‫‪B = ( x + 2 ) ( x + 1 ).‬‬‫ﻻ ﻴﻅﻬﺭ ﺍﻟﻌﺎﻤل ﺍﻟﻤﺸﺘﺭﻙ ﻨﻭﻅﻑ ‪C = x2 + 16x + 64‬‬‫ﺍﻟﻤﺘﻁﺎﺒﻘﺎﺕ ﺍﻟﺸﻬﻴﺭﺓ ‪C =x2 + 2(8x) 82‬‬‫‪C = ( x + 8 )2.‬‬ ‫‪(a+b)2=a2 +2ab +b2‬‬‫‪D = x2 – 9‬‬‫‪D = x2 –3a2‬‬‫)‪D =(x – 3)(x +3‬‬

‫‪ .4‬ﺍﻟﻤﻌﺎﺩﻻﺕ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭل ﻭﺍﺤﺩ ‪:‬‬ ‫ﺍﻟﺨﻭﺍﺹ‪:‬‬ ‫‪ A , B , C‬ﺃﻋﺩﺍﺩ‪.‬‬‫* ﻜل ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭل ﻭﺍﺤﺩ ‪X‬‬ ‫ﻴﻤﻜﻥ ﺘﺤﻭﻴﻠﻬﺎ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺸﻜل‪AX = B‬‬‫=‪X‬‬ ‫‪B‬‬ ‫* ﺇﺫﺍ ﻜﺎﻥ ‪ A ≠ 0‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ AX=B‬ﻫﻭ‬ ‫‪A‬‬ ‫ﻁﺭﻴﻘﺔ ﺍﻟﺤل‪:‬‬ ‫ﻟﺘﺤﻭﻴل ﻤﻌﺎﺩﻟﺔ ﺇﻟﻰ ﺍﻟﺸﻜل ‪ AX = B‬ﻨﺠﻤﻊ‬ ‫ﻭﺤﻴﺩﺍﺕ ﺍﻟﺤﺩ ﺍﻟﺘﻲ ﺘﺸﻤل ‪ X‬ﻋﻠﻰ ﺍﻟﻴﺴﺎﺭ ﻟﻠﻤﺴﺎﻭﺍﺓ‬ ‫ﻭ ﻭﺤﻴﺩﺍﺕ ﺍﻟﺤﺩ ﺍﻷﺨﺭﻯ ﻋﻠﻰ ﻴﻤﻴﻥ ﺍﻟﻤﺴﺎﻭﺍﺓ ‪.‬‬ ‫ﻤﺜﺎل)‪ : (1‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‪5X + 2 = 2X + 3 :‬‬ ‫ﻨﻁﺭﺡ –‪ 2‬ﻤﻥ ﻁﺭﻓﻲ ﺍﻟﻤﺴﺎﻭﺍﺓ‪5X+2-2 = 2X+3-2 :‬‬ ‫ﻨﻁﺭﺡ ‪ 2X‬ﻤﻥ ﻁﺭﻓﻲ ﺍﻟﻤﺴﺎﻭﺍﺓ‪5X=2X+1 :‬‬ ‫‪5X-2X=2X-2X+1‬‬ ‫‪3X = 1‬‬ ‫‪1‬‬ ‫×‬ ‫= ‪3X‬‬ ‫‪1‬‬ ‫×‬ ‫‪1‬‬ ‫‪:‬‬ ‫‪1‬‬ ‫ﻓﻲ‬ ‫ﺍﻟﻤﺴﺎﻭﺍﺓ‬ ‫ﻁﺭﻓﻲ‬ ‫ﻨﻀﺭﺏ‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫=‪X‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫= ‪ X‬ﻫﻲ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺩﺩﻴﺔ ﻟﻠﻤﺠﻬﻭل ‪ X‬ﺍﻟﺘﻲ ﻤﻥ ﺍﺠﻠﻬﺎ ﺘﺘﺤﻘﻕ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪.‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫ﻤﺜﺎل)‪: (2‬‬ ‫‪5‬‬ ‫‪X‬‬ ‫=‬ ‫‪4‬‬ ‫‪:‬‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ‫ﺤل‬ ‫‪3‬‬ ‫ﻨﻀﺭﺏ ﻁﺭﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻓﻲ ﻤﻘﻠﻭﺏ‬ ‫‪3‬‬ ‫×‬ ‫‪5‬‬ ‫=‪X‬‬ ‫‪3‬‬ ‫×‬ ‫‪4‬‬ ‫‪5‬‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪X‬‬ ‫=‬ ‫‪12‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫×‬ ‫‪12‬‬ ‫‪=4‬‬ ‫ﻨﺘﺤﻘﻕ‪:‬‬ ‫‪3‬‬ ‫‪5‬‬

‫ﻫﻭ ﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ‪.‬‬ ‫‪12‬‬ ‫ﻨﺴﺘﻨﺘﺞ ﺃﻥ‪:‬‬ ‫‪5‬‬ ‫‪ 1. 4‬ﻤﻌﺎﺩﻻﺕ ﻤﻥ ﺍﻟﺸﻜل ‪: (ax + b)(cx + d)=0‬‬ ‫ﺨﻭﺍﺹ‪:‬‬ ‫ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩﺍﻥ‪a;b:‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ‪ : a × b = 0‬ﻓﺈﻥ‪ a=0 :‬ﺃﻭ ‪b=0‬‬ ‫ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ (ax + b)(cx + d) = 0‬ﻫﻤﺎ ﺤﻠﻭل ﻜل ﻤﻥ‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ‪ ax + b = 0‬ﻭ ‪cx + d = 0‬‬ ‫ﻤﺜﺎل‪:‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‪(2x – 7)(-8x – 9) = 0 :‬‬ ‫* ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‪2x-7 = 0 :‬‬ ‫=‪x‬‬ ‫‪7‬‬ ‫ﺃﻱ‬ ‫‪2x = 7‬‬ ‫‪2‬‬ ‫* ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‪-8x-9 = 0 :‬‬ ‫=‪x‬‬ ‫‪9‬‬ ‫‪ -8x = 9‬ﺃﻱ‬ ‫‪−8‬‬‫‪.‬‬ ‫‪9‬‬ ‫ﻭ‬ ‫‪7‬‬ ‫ﻫﻤﺎ‪:‬‬ ‫ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ‪(2x-7)(-8x-9) = 0 :‬‬ ‫ﻭ ﻤﻨﻪ‪:‬‬ ‫‪−8‬‬ ‫‪2‬‬ ‫‪ .5‬ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭﻟﻴﻥ‪:‬‬ ‫ﻤﺜﺎل‪3x + 2y = -1 …(1)… :‬‬ ‫…)‪4x + y = 2 …(2‬‬ ‫ﺇﻨﻬﺎ ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻷﻭﻟﻰ ﺒﻤﺠﻬﻭﻟﻴﻥ ﺠﺒﺭﻴﻴﻥ ‪ Y‬ﻭ ‪x‬‬ ‫ﻤﻥ ﺍﺠل ‪ x = 1‬ﻭ ‪ y =-2‬ﺍﻟﻤﺴﺎﻭﺘﻴﻥ)‪ (1‬ﻭ )‪ (2‬ﻤﺤﻘﻘﺘﺎﻥ‬ ‫‪...(1)... 3x1 + 2(-2) = 3-4 = -1‬‬ ‫‪...(2)... 4x1 + (-2) = 4-2 = 2‬‬ ‫* ﻨﻘﻭل ﺍﻥ ﺍﻟﺜﻨﺎﺌﻴﺔ)‪ (-2 ، 1‬ﻟﻴﺴﺕ ﺤﻼ ﻟﻠﺠﻤﻠﺔ‪.‬‬ ‫* ﺒﻴﻨﻤﺎ ﺍﻟﺜﻨﺎﺌﻴﺔ )‪ (1 ،-2‬ﺤﻼ ﻟﻠﺠﻤﻠﺔ‪.‬‬ ‫ﻤﻼﺤﻅﺔ‪ :‬ﻓﻲ ﺜﻨﺎﺌﻴﺔ ﻋﺩﺩﻴﺔ ﺍﻟﺘﺭﺘﻴﺏ ﻫﺎﻡ‪.‬‬

‫‪ 1. 5‬ﺤل ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ‪:‬‬ ‫• ﻁﺭﻕ ﺤل ﺠﻤﻠﺔ ﻤﻌﺎﺩﻟﺘﻴﻥ‪:‬‬ ‫‪ -‬ﻁﺭﻴﻘﺔ ﺍﻟﺤل ﺒﺎﻟﺘﻌﻭﻴﺽ‪:‬‬‫ﺘﻬﺩﻑ ﻁﺭﻴﻘﺔ ﺍﻟﺤل ﺒﺎﻟﺘﻌﻭﻴﺽ ﺇﻟﻰ ﺍﺴﺘﺨﺭﺍﺝ ﺃﺤﺩ ﺍﻟﻤﺠﻬﻭﻟﻴﻥ ﻤﻥ ﺇﺤﺩﻯ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ ﺜﻡ‬ ‫ﺍﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻷﺨﺭﻯ‪.‬‬ ‫ﻤﺜﺎل ‪ :‬ﺤل ﺍﻟﺠﻤﻠﺔ ﺍﻵﺘﻴﺔ ‪:‬‬ ‫…)‪3x + 2y = -1 …(1‬‬ ‫…)‪4x + y = 2 …(2‬‬ ‫ﺍﻟﻤﻌﺎﺩﻟﺔ‪...(2)...‬ﻤﺜﻼ ﺘﺴﻤﺢ ﺒﻜﺘﺎﺒﺔ ‪y = 2-4x :‬‬ ‫ﻨﻌﻭﺽ ‪ y = 2-4x‬ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ‪...(1)...‬ﻓﻨﺠﺩ‪:‬‬ ‫‪3x + 2(2-4x) = -1‬‬ ‫‪3x + 4-8x = -1‬‬ ‫‪ -5x = -5‬ﺃﻱ‪x = 1 :‬‬ ‫ﻨﻌﻭﺽ ‪ x = 1‬ﻓﻲ ‪y = 2-4x‬‬ ‫ﻨﺠﺩ‪y = 2-4× 1 :‬‬ ‫‪y = 2-4‬‬ ‫‪Y = -2‬‬ ‫‪ -‬ﻁﺭﻴﻘﺔ ﺍﻟﺤل ﺒﺎﻟﺠﻤﻊ‪:‬‬ ‫ﺘﻬﺩﻑ ﻁﺭﻴﻘﺔ ﺍﻟﺤل ﺒﺎﻟﺠﻤﻊ ﺇﻟﻰ ﺠﻌل ﻤﻌﺎﻤﻠﻲ ﺱ ﺃﻭ ﻉ ﻤﺘﻌﺎﻜﺴﻴﻥ ﺜﻡ ﺘﺠﻤﻊ ﻁﺭﻑ ﻤﻊ ﻁﺭﻑ‬ ‫ﻤﺜﺎل ‪ :‬ﻨﺘﻌﺎﻤل ﻤﻊ ﻨﻔﺱ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫…)‪1 × 3x+2y = -1 …(1‬‬ ‫)‪(-2) × 4x+y = 2 …(2‬‬ ‫…)‪3x+2y = -1 …(‘1‬‬ ‫…)‪-8x-2y = -4 …(‘2‬‬ ‫‪3x-8x+2y-2y = -1-4‬‬