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Snally
@Snally
July 2022
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3*4^х + 2*9^х - 5*6^х < 0
2^х+1 + 4^х ≤ 80
Помогите, пожалуйста, решить неравенства
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pomoshnik09
3*2^(2x)+2*3^(2x)-5*2^x*3^x<0 разделим на 2^x*3^x>0
3*(2/3)^x+2*(3/2)^x -5<0
пусть 2/3)^x=t t>0
3t-2/t-5<0. 3t^2-5t+2<0 D=5^2-4*3*2=25-24=1 t1=5+1)/6=1 t2=5-1)/6=
=2/3
2/3<t<1
2/3<(2/3)^x<1
(2/3)^1<2/3)^x<(2/3)^0 убывающая функция
0<x<1 x∈(0. 1)
2) 2*2^x+2^(2x)-80≤0 пусть 2^x=t. тогда
t²+2t-80≤0 D=324 t1=-10 t2=8
2^x≤8 2^x≤2³ x≤3 x∈(-∞. 3] наверно так
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Answers & Comments
3*(2/3)^x+2*(3/2)^x -5<0
пусть 2/3)^x=t t>0
3t-2/t-5<0. 3t^2-5t+2<0 D=5^2-4*3*2=25-24=1 t1=5+1)/6=1 t2=5-1)/6=
=2/3
2/3<t<1
2/3<(2/3)^x<1
(2/3)^1<2/3)^x<(2/3)^0 убывающая функция
0<x<1 x∈(0. 1)
2) 2*2^x+2^(2x)-80≤0 пусть 2^x=t. тогда
t²+2t-80≤0 D=324 t1=-10 t2=8
2^x≤8 2^x≤2³ x≤3 x∈(-∞. 3] наверно так