1. Solve the equation: $\sqrt{x^{2}-4} \cdot|x|+\sqrt{x^{2}-4} \cdot x=0$.
Solution. Since $\sqrt{x^{2}-4} \geq 0$, we consider two cases.

First case $\sqrt{x^{2}-4}=0$. Hence, $x= \pm 2-$ roots.

Second case. $\sqrt{x^{2}-4}>0$, hence $|x|>2$. Then the given equation can be rewritten as follows: $|x|+x=0$. From this, it follows that the solution to the last equation is the interval $x \leq 0$. Taking into account the first case and the condition $|x|>2$, we get the answer: $x \in(-\infty ;-2] \cup\{2\}$.

Answer: $x \in(-\infty ;-2] \cup\{2\}$