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ghagha0
@ghagha0
August 2022
1
16
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16^x-12^x-2*9^x<0
как решать?
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sedinalana
Verified answer
4^2x-4^x*3^x-2*3^2x<0/3^2x
(4/3)^2x-(4/3)^x-2<0
(4/3)^x=a
a²-a-2<0
a1+a2=1 U a1*a2=-2
a1=-1 U a2=2
-1<a<2
-1<(4/3)^x<2
x<log(4/3)2
x∈(-∞;log(4/3)2)
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Answers & Comments
Verified answer
4^2x-4^x*3^x-2*3^2x<0/3^2x(4/3)^2x-(4/3)^x-2<0
(4/3)^x=a
a²-a-2<0
a1+a2=1 U a1*a2=-2
a1=-1 U a2=2
-1<a<2
-1<(4/3)^x<2
x<log(4/3)2
x∈(-∞;log(4/3)2)