Срочно

$\log_{2} (\frac{8}{x})-\frac{10}{\log_{2}(16x)}\geq0 \\ \\ \left \{ {{\frac{8}{x}\ \textgreater \ 0} \atop {16x\ \textgreater \ 0}} \right. =\ \textgreater \ x\ \textgreater \ 0 \\\\ \log_{2} 8- \log_{2} x-\frac{10}{\log_{2}(16)+\log_{2}x} \geq0 \\ \\ 3-\log_{2}x-\frac{10}{4+\log_{2}x}\geq0 \\ \\ \[\log_{2}x=t =\ \textgreater \ 3-t-\frac{10}{t+4}\geq0 \\ \\ \frac{3*(t+4)-t*(t+4)-10}{t+4}\geq0 \\ \\ \frac{3t+12-t^2-4t-10}{t+4}\geq0 \\ \\ \frac{-t^2-t+2}{t+4}\geq0 \ | : (-1) \\ \\ \frac{t^2+t-2}{t+4}\leq 0 \\ \frac{(t-1)(t+2)}{t+4} \leq 0 \\ \\$
<смотрите приложенный скриншот>
$\left \{ {{t\ \ \textless \ \ -4} \atop {-2 \ \leq \ t \ \leq \ 1}} \right. =\ \textgreater \ \left \{ {{\log_{2} x\ \ \textless \ \ -4} \atop {-2 \ \leq \ \log_{2} x \ \leq \ 1}} \right. =\ \textgreater \ \left \{ {{{x\ \ \textless \ \ 2^{-4}} \atop {2^{-2} \ \leq \ x \ \leq \ 2^1}} \right. =\ \textgreater \ \\ \\ \left \{ {{x \ \ \textless \ \ \frac{1}{16}} \atop {\frac{1}{4} \ \leq \ x \ \leq \ 2}} \right. \\ \\ x \ \textgreater \ 0 =\ \textgreater \ x \in (0; \frac{1}{16}) \ \cup \ [\frac{1}{4}; 2].$