1. Solve the inequality $\log _{\frac{x^{2}-6}{4 x-10}}\left(\frac{\left(x^{2}-6\right)(4 x-10)}{9}\right) \geq 1$.
Answer: $x \in\left[\frac{7}{4} ; 2\right) \cup(2 ; \sqrt{6}) \cup\left[\frac{13}{4} ;+\infty\right)$.

Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-6}{4 x-10}>0, \frac{x^{2}-6}{4 x-10} \neq 1$ (then the logarithmic expression is also positive), from which $x \in(-\sqrt{6} ; \sqrt{6}) \cup\left(\frac{5}{2} ;+\infty\right), x \neq 2$.

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

$$
\begin{aligned}
& \left.\left(\frac{x^{2}-6}{4 x-10}-1\right)\left(\frac{\left(x^{2}-6\right)(4 x-10)}{9}-\frac{x^{2}-6}{4 x-10}\right) \geq 0 \Leftrightarrow \frac{x^{2}-4 x+4}{4 x-10} \cdot \frac{\left(x^{2}-6\right)\left((4 x-10)^{2}-9\right]}{9(4 x-10)}\right]_{\geq 0} \Leftrightarrow \\
& (x-2)^{2}(x-\sqrt{6})(x+\sqrt{6})(4 x-13)(4 x-7) \geq 0 \Leftrightarrow x \in(-\infty ;-\sqrt{6}] \cup\left[\frac{7}{4} ; \sqrt{6}\right] \cup\left[\frac{13}{4} ;+\infty\right)
\end{aligned}
$$

Considering the domain, we get $x \in\left[\frac{7}{4} ; 2\right) \cup(2 ; \sqrt{6}) \cup\left[\frac{13}{4} ;+\infty\right)$.