1. Solve the inequality $\log _{\frac{x^{2}-3}{6 x-12}}\left(\frac{\left(x^{2}-3\right)(6 x-12)}{25}\right) \geq 1$.
Answer: $x \in\left[\frac{7}{6} ; \sqrt{3}\right) \cup\left[\frac{17}{6} ; 3\right) \cup(3 ;+\infty)$.

Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-3}{6 x-12}>0, \frac{x^{2}-3}{6 x-12} \neq 1$ (then the logarithmic expression is also positive), from which $x \in(-\sqrt{3} ; \sqrt{3}) \cup(2 ;+\infty), x \neq 3$.

Within the domain, the given inequality is equivalent to each of the following inequalities:

$$
\begin{aligned}
& \left(\frac{x^{2}-3}{6 x-12}-1\right)\left(\frac{\left(x^{2}-3\right)(6 x-12)}{25}-\frac{x^{2}-3}{6 x-12}\right) \geq 0 \Leftrightarrow \frac{x^{2}-6 x+9}{6 x-12} \cdot \frac{\left.\left(x^{2}-3\right)(6 x-12)^{2}-25\right]}{25(6 x-12)} \geq 0 \Leftrightarrow \\
& (x-3)^{2}(x-\sqrt{3})(x+\sqrt{3})(6 x-17)(6 x-7) \geq 0 \Leftrightarrow x \in(-\infty ;-\sqrt{3}] \cup\left[\frac{7}{6} ; \sqrt{3}\right] \cup\left[\frac{17}{6} ;+\infty\right)
\end{aligned}
$$

Considering the domain, we get $x \in\left[\frac{7}{6} ; \sqrt{3}\right) \cup\left[\frac{17}{6} ; 3\right) \cup(3 ;+\infty)$.