1. Solve the inequality $\sqrt{x^{2}-25} \cdot \sqrt{-2 x-1} \leq x^{2}-25$.
Answer. $x \in(-\infty ;-6] \cup\{-5\}$.

Solution. The domain of definition of the given inequality is the set $x \in(-\infty ;-5]$. Consider two cases.

a) When $x=-5$, the inequality is satisfied (we get $0=0$).

b) When $x<-5$, we divide both sides of the inequality by the positive number $\sqrt{x^{2}-25}$ and obtain $\sqrt{-2 x-1} \leq \sqrt{x^{2}-25}$; then $-2 x-1 \leq x^{2}-25, x^{2}+2 x-24 \geq 0, x \in(-\infty ;-6] \cup[4 ;+\infty)$. Taking into account the condition, we get $x \in(-\infty ;-6]$.

Combining the results, we find $x \in(-\infty ;-6] \cup\{-5\}$.