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August 2022
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Решить методом замены переменной
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dasdasfa
A) 3*3^x *2 *2^(-x)+3^x*2^(-x)≤10,5
6*(3/2)^x +(3/2)^x≤10,5
t=(3/2)^x; 6t+t≤10,5
7t≤10,5; t≤10,5:7; t≤1,5
(3/2)^x≤1,5
1,5^x≤1,5
0<x≤1
2 votes
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Answers & Comments
6*(3/2)^x +(3/2)^x≤10,5
t=(3/2)^x; 6t+t≤10,5
7t≤10,5; t≤10,5:7; t≤1,5
(3/2)^x≤1,5
1,5^x≤1,5
0<x≤1