1. Solve the inequality $\sqrt{x^{2}-4} \cdot \sqrt{2 x-1} \leq x^{2}-4$.
Answer. $x \in\{2\} \cup[3 ;+\infty)$.

Solution. The domain of definition of the given inequality is the set $x \in[2 ;+\infty)$. Consider two cases.

a) For $x=2$, the inequality is satisfied (we get $0=0$).

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

Combining the results, we find $x \in\{2\} \cup[3 ;+\infty)$.