1. Solve the inequality $\log _{\frac{x^{2}-5}{2 x-6}}\left(\frac{\left(x^{2}-5\right)(2 x-6)}{25}\right) \geq 1$.
Answer: $x \in\left[\frac{1}{2} ; 1\right) \cup(1 ; \sqrt{5}) \cup\left[\frac{11}{2} ;+\infty\right)$.

Solution. The domain of the inequality is defined by the conditions $\frac{x^{2}-5}{2 x-6}>0, \frac{x^{2}-5}{2 x-6} \neq 1$ (then the logarithmic expression is also positive), from which $x \in(-\sqrt{5} ; \sqrt{5}) \cup(3 ;+\infty), x \neq 1$.

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

$$
\begin{aligned}
& \left(\frac{x^{2}-5}{2 x-6}-1\right)\left(\frac{\left(x^{2}-5\right)(2 x-6)}{25}-\frac{x^{2}-5}{2 x-6}\right) \geq 0 \Leftrightarrow \frac{x^{2}-2 x+1}{2 x-6} \cdot \frac{\left(x^{2}-5\right)}{25(2 x-6)}\left[(2 x-6)^{2}-25\right] \geq 0 \Leftrightarrow \\
& (x-1)^{2}(x-\sqrt{5})(x+\sqrt{5})(2 x-11)(2 x-1) \geq 0 \Leftrightarrow x \in(-\infty ;-\sqrt{5}] \cup\left[\frac{1}{2} ; \sqrt{5}\right] \cup\left[\frac{11}{2} ;+\infty\right)
\end{aligned}
$$

Considering the domain, we get $x \in\left[\frac{1}{2} ; 1\right) \cup(1 ; \sqrt{5}) \cup\left[\frac{11}{2} ;+\infty\right)$.