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evagoss
@evagoss
July 2022
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Решить неравенства :(((
[tex] (\frac{3}{4})^{x} \ \textgreater \ 1
3^{x+2} + 3^{x+1} \ \textless \ 28[/tex]
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kirichekov
Verified answer
(3/4)^x>1
(3/4)^x>(3/4)⁰
a=3/4. 0<3/4<1. знак неравенства меняем
x<0
3^(x+2)+3^(x+1)<28
3^x*3²+3^x*3¹<28
3^x*(9+3)<28
3^x<28:12
3^x<7/3
log₃3^x<log₃(7/3)
x*log₃3<log₃(7/3)
x<log₃(7/3)
если бы условие было:
3^(x+2)+3^(x-1)<28
3^x*3²+3^x*3⁻¹<28
3^x*(9+1/3)<28
3^x<28:(28/3)
3^x<3
3^x<3¹. a=3, 3>1 знак неравенства не меняем
x<1
2 votes
Thanks 1
evagoss
спасибо!
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Answers & Comments
Verified answer
(3/4)^x>1(3/4)^x>(3/4)⁰
a=3/4. 0<3/4<1. знак неравенства меняем
x<0
3^(x+2)+3^(x+1)<28
3^x*3²+3^x*3¹<28
3^x*(9+3)<28
3^x<28:12
3^x<7/3
log₃3^x<log₃(7/3)
x*log₃3<log₃(7/3)
x<log₃(7/3)
если бы условие было:
3^(x+2)+3^(x-1)<28
3^x*3²+3^x*3⁻¹<28
3^x*(9+1/3)<28
3^x<28:(28/3)
3^x<3
3^x<3¹. a=3, 3>1 знак неравенства не меняем
x<1