1. Solve the inequality $\sqrt{x^{2}-16} \cdot \sqrt{2 x-1} \leq x^{2}-16$.
Answer. $x \in\{4\} \cup[5 ;+\infty)$.

Solution. The domain of definition of the given inequality is the set $x \in[4 ;+\infty)$. Consider two cases.

a) When $x=4$, the inequality is satisfied (we get $0=0$).

b) When $x>4$, divide both sides of the inequality by the positive number $\sqrt{x^{2}-16}$ and we get $\sqrt{2 x-1} \leq \sqrt{x^{2}-16}$; then $2 x-1 \leq x^{2}-16, x^{2}-2 x-15 \geq 0, x \in(-\infty ;-3] \cup[5 ;+\infty)$. Considering the condition, we get $x \in[5 ;+\infty)$.

Combining the results, we find $x \in\{4\} \cup[5 ;+\infty)$.