1. Solve the inequality $\frac{\log _{2}\left(x^{6}\right) \cdot \log _{\frac{1}{2}}\left(x^{2}\right)-\log _{\frac{1}{2}}\left(x^{6}\right)-8 \log _{2}\left(x^{2}\right)+2}{8+\left(\log _{\frac{1}{2}}\left(x^{2}\right)\right)^{3}} \leq 0$.
Answer. $x \in(-2 ;-\sqrt[6]{2}] \cup\left[-\frac{1}{2} ; 0\right) \cup\left(0 ; \frac{1}{2}\right] \cup[\sqrt[6]{2} ; 2)$.

Solution. The given inequality is equivalent to the following:

$$
\frac{3 \log _{2} x^{2} \cdot\left(-\log _{2} x^{2}\right)+3 \log _{2} x^{2}-8 \log _{2} x^{2}+2}{8-\log _{2}^{3} x^{2}} \leq 0
$$

After the substitution $\log _{2} x^{2}=t$, the inequality becomes $\frac{-3 t^{2}-5 t+2}{8-t^{3}} \leq 0$, from which $\frac{(2+t)(1-3 t)}{2-t} \leq 0$,

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

We find the values of $x$.

For $t \leq-2$ we get $0<x^{2} \leq \frac{1}{4} \Leftrightarrow x \in\left[-\frac{1}{2} ; 0\right) \cup\left(0 ; \frac{1}{2}\right]$.

For $\frac{1}{3} \leq t<2$ we get $\sqrt[3]{2} \leq x^{2}<4 \Leftrightarrow x \in(-2 ;-\sqrt[6]{2}] \cup[\sqrt[6]{2} ; 2)$.