ﺍﺧﺘﺒﺎﺭﺗﺸﺨﻴﺼﻲﻟﻠﺼﻒﺍﻷﻭﻝﺛﺎﻧﻮﻱﺍﻟﻌﻠﻤﻲ ************************************************************************************* ********* (1ﺑﺎﻗﻲﻗﺴﻤﺔﺍﻻﻗﺘﺮﺍﻥﻕ)ﺱ(=ﺱ2-3ﺱ4+2ﻋﻠﻰﺍﻻﻗﺘﺮﺍﻥﻩـ)ﺱ(= ﺱ3-ﻳﺴﺎﻭﻱ ﺩ(13- ﺝ(13 ﺏ(3- ﺃ(3 (2ﺍﻟﺼﻴﻐﺔﺍﻟﻤﻜﺎﻓﺌﺔﻟﻼﻗﺘﺮﺍﻥﺍﻟﻨﺴﺒﻲ ﺱ +2ﺱ6-ﻳﺴﺎﻭﻱ ﺩ( ﺱ–2 ﺝ( ﺱ–2 ﺏ(1 ﺃ( ﺱ2+ ﺝ() (∞ ،2)،[3-،∞ -ﺩ(][2،3- (3ﻣﺠﻤﻮﻋﺔﺣﻞﺍﻟﻤﺘﺒﺎﻳﻨﺔ ﺱ -2ﺱ–6 ﺃ() (∞ ،3]،[2-،∞ -ﺏ(][3،2- (4ﺍﺫﺍﻛﺎﻥ ﺱ = ﺃ +ﺏ ﻓﺈﻥﻗﻴﻤﺔﺃ،ﺏﻋﻠﻰﺍﻟﺘﺮﺗﻴﺐﻫﻲ ﺩ(1-،1 ﺝ(1،1- ﺏ(1،0 ﺃ(0،1 (5ﻣﺠﺎﻝﺍﻻﻗﺘﺮﺍﻥﻉ)ﺱ(= ﺱ 1+ﻳﺴﺎﻭﻱ ﺩ()(1-،∞ - ﺝ()(∞ ،1- ﺃ(] (∞ ،1-ﺏ()[1-،∞ -
(6ﺍﻟﺮﺳﻤﺔﺍﻟﻤﺠﺎﻭﺭﺓﻫﻲﺭﺳﻤﺔﺍﻻﻗﺘﺮﺍﻥﻕ)ﺱ(= ﺏ( ﺱ1+ ﺃ( ﺱ ﺩ( ﺱ–1 ﺝ( ﺱ (7ﻣﺠﻤﻮﻋﺔﺍﻟﺤﻞﻟﻠﻤﻌﺎﺩﻟﺔ lﺱ–5= 1ﻫﻲ ﺩ(}{6-،4 ﺝ(}{6-،4- ﺏ(}{6،4- ﺃ(}{6،4 (8ﺍﺫﺍﻛﺎﻥﻕ)ﺱ(=2ﺱ،3+ﻩـ)ﺱ(= ﺱ1–2ﻓﺈﻥ)ﻕﻩـ()ﺱ(ﻳﺴﺎﻭﻱ ﺩ( 2ﺱ1+2 ﺝ(4ﺱ 12+2ﺱ8+ ﺃ(2ﺱ 3+2ﺏ(2ﺱ2+2 (9ﺍﺫﺍﻛﺎﻥﻕ)ﺱ(= 3-2ﺱ ﻓﺈﻥﻕ)1-ﺱ(= ﺩ( ﺱ2+ ﺝ( 1–2ﺱ ﺏ( –2ﺱ ﺃ( 1ﺱ 2+ (10ﺃﺣﺪﺍﻻﻗﺘﺮﺍﻧﺎﺕﺍﻟﺘﺎﻟﻴﺔﻫﻮﺍﻗﺘﺮﺍﻥﻭﺍﺣﺪﻟﻮﺍﺣﺪ
ﺩ(ﻝ)ﺱ(= ﺝ(ﻙ)ﺱ(= ﺱ–1 ﺏ(ﻩـ)ﺱ(= ﺱ1+ ﺃ(ﻕ)ٍﺱ(= ﺱ ﺱ2
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