1. Solve the inequality $\sqrt{x^{2}-9} \cdot \sqrt{-2 x-1} \leq x^{2}-9$.
Answer. $x \in(-\infty ;-4] \cup\{-3\}$.

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

a) When $x=-3$, the inequality is satisfied (we get $0=0$).

b) When $x<-3$, we divide both sides of the inequality by the positive number $\sqrt{x^{2}-9}$ and obtain $\sqrt{-2 x-1} \leq \sqrt{x^{2}-9}$; then $-2 x-1 \leq x^{2}-9, x^{2}+2 x-8 \geq 0, x \in(-\infty ;-4] \cup[2 ;+\infty)$. Taking into account the condition, we get $x \in(-\infty ;-4]$.

Combining the results, we find $x \in(-\infty ;-4] \cup\{-3\}$.