מתמטיקה - 4 יח"ל - כיתה י"ב - חלק א'
ﺟﺑﺭ ﻭﻣﻘﺩّﻣﺔ ﻟﻠﺗّﺣﻠﻳﻝ ﺍﻟﺭّﻳﺎﺿﻲ ﻟﻠﺩّﻭﺍﻝ ﺍﻷﺳّﻳّﺔ ﻭﺍﻟﻠّﻭﻏﺭﻳﺛﻣﻳّﺔ
- 75 -
ﺍﻟﻔﺻﻝ :7 ﻣﻌﺎﺩﻻﺕ ﻟﻭﻏﺭﻳﺛﻣﻳّﺔ ﻭﻣﺗﺑﺎﻳﻧﺎﺕ ﻟﻭﻏﺭﻳﺛﻣﻳّﺔ
2
)
(3) ﺣﻠّﻮﺍ ﺍﻟﻤﺘﺒﺎﻳﻨﺔ :
2 log (
) log (
1 2
x
x
2
اﻟﺤﻞّ: ﻧﺤﻮّﻝ ﺍﻟﻄّﺮﻑ ﺍﻷﻳﻤﻦ ﺇﻟﻰ ﻛﺘﺎﺑﺔ ﺑﻮﺍﺳﻄﺔ ﻟﻮﻏﺮﻳﺜﻢ ﻟﻸﺳﺎﺱ . 2 ﺗﺬﻛﻴﺮ :
b a log a b
log
2
ﺃﻱ :
log
2 2
4 2
2
2
ﻭﻟﺬﺍ، ﻧﻜﺘﺐ ﻓﻲ ﺍﻟﻄّﺮﻑ ﺍﻷﻳﻤﻦ 2 4 log ﺑﺪَﻝَ ﺍﻟﻌﺪﺩ . 2 ﻧﺴﺘﻌﻤﻞ ﻓﻲ ﺍﻟﻄّﺮﻑ ﺍﻷﻳﺴﺮ ﻗﺎﻧﻮﻥ ﺣﺎﺻﻞ ﺍﻟﻀّﺮﺏ ﻓﻲ ﺍﻟﻠّﻮﻏﺮﻳﺜﻤﺎﺕ، ﻧﺤﺼﻞ ﻋﻠﻰ: x x _ i ﻣﻦ ﺍﻟﻤﺘﺒﺎﻳﻨﺔ ﺍﻟﻠّﻮﻏﺮﻳﺜﻤﻴّﺔ ) ﺃﺳﺎﺱ ﺍﻟﻠّﻮﻏﺮﻳﺜﻢ ﺃﻛﺒﺮ ﻣﻦ ( 1 ﻭﻣﻦ ﻣﺠﺎﻝ ﺍﻟﺘّﻌﻮﻳﺾ، ﻧﺤﺼﻞ ﻋﻠﻰ ﻫﻴﺌﺔ ﺍﻟﻤﺘﺒﺎﻳﻨﺎﺕ : 2 2 1 4 )( ) log
2 log (
2 1 4 2 0 1 0 ( )( ) x x x x x x
3 x
2
6 0
2
x x
3
x
2
ﻭﺃﻳﻀًﺎ
2
1
ﻧﺤﻞ ﻫﻴﺌﺔ ﺍﻟﻤﺘﺒﺎﻳﻨﺎﺕ ﺑﻤﺴﺎﻋﺪﺓ ﺍﻟﺮّﺳﻢ ﻋﻠﻰ ﻣﺤﻮﺭ ﺍﻷﻋﺪﺍﺩ، ﺃﻱ ﺃﻥ ﺍﻟﺤﻞ ﻫﻮ 1 2 x .
x
3
2
1
2
ﺗﻤﺎﺭﻳﻦ ﻟﻠﻌﻤﻞ ﺍﻟﺬّﺍﺗﻲ
ﺣﻠّﻮﺍ ﺍﻟﻤﺘﺒﺎﻳﻨﺎﺕ ﻓﻲ ﺍﻟﺘّﻤﺎﺭﻳﻦ (10) – (1) . (1) 4 4 2 1 3 2 ) log ( ) x x
log (
1 log (
1 ) log (
)
(2)
5 4 x
3 6 x
1 _ i 3 x
4
4
1
2
log (
) log (
)
log
3 log (
)
(4)
(3)
7
5
x
x
x
3
2 3 log ( ) log x x
3 x
3 log (
) log
(6)
(5)
5
3
x
x
log
log
ln (
) ln
(8)
(7)
10
3 1
2
0.4
0.4
2 1 ln ln ( ) x x
) ln ( x
ln (
)
(10)
(9)
2 5 x
3 3
ﺣﻠّﻮﺍ ﺍﻟﻤﺘﺒﺎﻳﻨﺎﺕ ﻓﻲ ﺍﻟﺘّﻤﺎﺭﻳﻦ (31) – (11) . (11) 4 0 log x
) x
x
3 log (
log
(13)
(12)
1 2
1
2
x
5 2 1 2 log ( ) x
) x
3 log (
log
(16)
(15)
(14)
2 1
2
1 2
1 2 2 1 1 log ( ) x 3 2 1 1 log ( ) x
x
) x
log
log (
(19)
(18)
(17)
1
2 2
1 4
1 3
1 5 2 5 2 log ( ) x
) x
6 log (
(22)
(21)
(20)
4 0
ln x
) x
) x
2 log (
0.5 log (
(25)
(24)
(23)
4 1 0
4
3
2
0
) x
2 1 0 ln x
ln x
ln (
(28)
(27)
(26)
3
0
2
1 0
2 x x _ i 3 2 1 )( )
2 ) x x 3
log (
6 log (
(30)
(29)
1
1 4
2 ) x x 6 9 2 0
log (
(31)
1 3
© ﺟﻣﻳﻊ ﺍﻟﺣﻘﻭﻕ ﻣﺣﻔﻭﻅﺔ ﻟﺟﺎﺑﻲ ﻳﻛﻭﺋﻳﻝ –
– ﺭﻳﺎﺿﻳّﺎﺕ ﻟﻁﻼّﺏ 4 ﻭﺣﺩﺍﺕ ﺗﻌﻠﻳﻣﻳّﺔ – ﺍﻟﺻّﻑ ﺍﻟﺛّﺎﻧﻲ ﻋﺷﺭ – ﺍﻟﻣﻧﻬﺞ ﺍﻟﺟﺩﻳﺩ
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