Question

Решите неравенство, задания прикреплены...

Answer 1

$1)\quad (0,2)^{\frac{6x-1}{3-x}}\ \textless \ (\frac{1}{5} )^2\\\\0,2= \frac{1}{5} \\\\ \frac{6x-1}{3-x}\ \textgreater \ 2 \; ,\; \; \frac{6x-1-2(3-x)}{3-x} \ \textgreater \ 0\; ,\; \; \frac{6x-1-6+2x}{3-x}\ \textgreater \ 0\; ,\; \; \frac{8x-7}{x-3} \ \textless \ 0\\\\+++( \frac{7}{8})---(3)+++\\\\x\in ( \frac{7}{8} ,3)\\\\2)\quad (\frac{1}{2} )^{\frac{x-1}{x+2}} \geq 4\\\\2^{\frac{-x+1}{x+2}} \geq 2^2\\\\ \frac{-x+1}{x+2} \geq 2\; ,\; \; \frac{-x+1-2(x+2)}{x+2} \geq 0\; ,\; \; \frac{-3x-3}{x+2} \geq 0\; ,\; \frac{3(x+1)}{x+2} \leq 0\\\\++(-2)---[-1\, ]+++$

$x\in (-2,-1\, ]$