1. Solve the inequality $\frac{125+\left(\log _{\frac{1}{2}}\left(x^{4}\right)\right)^{3}}{\log _{2}\left(x^{4}\right) \cdot \log _{2}\left(x^{2}\right)+6 \log _{\frac{1}{2}}\left(x^{4}\right)+17 \log _{2}\left(x^{2}\right)-3} \geq 0$.
Answer. $x \in[-2 \sqrt[4]{2} ;-\sqrt[4]{2}) \cup\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right) \cup(\sqrt[4]{2} ; 2 \sqrt[4]{2}]$.

Solution. The given inequality is equivalent to the following:

$$
\frac{125-\left(2 \log _{2}\left(x^{2}\right)\right)^{3}}{2 \log _{2}\left(x^{2}\right) \cdot \log _{2}\left(x^{2}\right)-12 \log _{2}\left(x^{2}\right)+17 \log _{5}\left(x^{2}\right)-3} \geq 0
$$

After the substitution $\log _{2} x^{2}=t \quad$ the inequality takes the form $\quad \frac{125-(2 t)^{3}}{2 t^{2}+5 t-3} \geq 0$, from which $\frac{2 t-5}{(t+3)(2 t-1)} \leq 0$,

$$
t \in(-\infty ;-3) \cup\left(\frac{1}{2} ; \frac{5}{2}\right]
$$

We find the values of $x$.

For $t<-3$ we get $0<x^{2}<\frac{1}{8} \Leftrightarrow x \in\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right)$.

For $\frac{1}{2}<t \leq \frac{5}{2}$ we get $\sqrt{2}<x^{2} \leq \sqrt{32} \Leftrightarrow x \in[-2 \sqrt[4]{2} ;-\sqrt[4]{2}) \cup(\sqrt[4]{2} ; 2 \sqrt[4]{2}]$.