1. Solve the inequality $\frac{64+\left(\log _{\frac{1}{5}}\left(x^{2}\right)\right)^{3}}{\log _{\frac{1}{5}}\left(x^{6}\right) \cdot \log _{5}\left(x^{2}\right)+5 \log _{5}\left(x^{6}\right)+14 \log _{\frac{1}{5}}\left(x^{2}\right)+2} \leq 0$.
Answer. $x \in[-25 ;-\sqrt{5}) \cup\left(-\frac{1}{\sqrt[3]{5}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[3]{5}}\right) \cup(\sqrt{5} ; 25]$.

Solution. The given inequality is equivalent to the following:

$$
\frac{64-\left(\log _{5}\left(x^{2}\right)\right)^{3}}{-3 \log _{5}\left(x^{2}\right) \cdot \log _{5}\left(x^{2}\right)+15 \log _{5}\left(x^{2}\right)-14 \log _{5}\left(x^{2}\right)+2} \leq 0
$$

After the substitution $\log _{5} x^{2}=t$, the inequality becomes $\frac{64-t^{3}}{-3 t^{2}+t+2} \leq 0$, from which $\frac{4-t}{(1-t)(2+3 t)} \leq 0$,

$$
t \in\left(-\infty ;-\frac{2}{3}\right) \cup(1 ; 4]
$$

We find the values of $x$.

For $t<-\frac{2}{3}$, we get $0<x^{2}<5^{-2 / 3} \Leftrightarrow x \in\left(-\frac{1}{\sqrt[3]{5}} ; 0\right) \cup\left(0 ; \frac{1}{\sqrt[3]{5}}\right)$.

For $1<t \leq 4$, we get $5<x^{2} \leq 625 \Leftrightarrow x \in[-25 ;-\sqrt{5}) \cup(\sqrt{5} ; 25]$.