Question

Решите неравенство :
(6^x-4*3^x):(x*2^x-5*2^x-4x+20)≤1:(x-5)

Answer 1

$\frac{6^x-4*3^x}{x*2^x-5*2^x-4x+20} \leq \frac{1}{x-5} \\ \\ \frac{2^x*3^x-4*3^x}{2^x(x-5)-4(x-5)} \leq \frac{1}{x-5} \\ \\ \frac{3^x(2^x-4)-(2^x-4)}{(x-5)(2^x-4)} \leq 0 \\ \\ \frac{(2^x-4)(3^x-1)}{(x-5)(2^x-4)} \leq 0 \\ \\ \frac{(2^x-2^2)(3^x-3^0)}{(x-5)(2^x-2^2)} \leq 0 \\ \\ \frac{(x-2)(x-0)}{(x-5)(x-2)} \leq 0$

++++[0]------(2)-------(5)+++++

x∈[0;2)U(2;5)