1. Solve the inequality $\log _{\frac{x^{2}-2}{2 x-3}}\left(\frac{\left(x^{2}-2\right)(2 x-3)}{4}\right) \geq 1$.
Answer: $x \in\left[\frac{1}{2} ; 1\right) \cup(1 ; \sqrt{2}) \cup\left[\frac{5}{2} ;+\infty\right)$.

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

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

$$
\begin{aligned}
& \left(\frac{x^{2}-2}{2 x-3}-1\right)\left(\frac{\left(x^{2}-2\right)(2 x-3)}{4}-\frac{x^{2}-2}{2 x-3}\right) \geq 0 \Leftrightarrow \frac{x^{2}-2 x+1}{2 x-3} \cdot \frac{\left.\left(x^{2}-2\right)(2 x-3)^{2}-4\right]}{4(2 x-3)} \geq 0 \Leftrightarrow \\
& (x-1)^{2}(x-\sqrt{2})(x+\sqrt{2})(2 x-5)(2 x-1) \geq 0 \Leftrightarrow x \in(-\infty ;-\sqrt{2}] \cup\left[\frac{1}{2} ; \sqrt{2}\right] \cup\left[\frac{5}{2} ;+\infty\right) .
\end{aligned}
$$

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