2.022. $\frac{\sqrt{(x+2)^{2}-8 x}}{\sqrt{x}-\frac{2}{\sqrt{x}}}$.

2.022. $\frac{\sqrt{(x+2)^{2}-8x}}{\sqrt{x}-\frac{2}{\sqrt{x}}}$.

## Solution.

Domain of definition: $0<x \neq 2$.

$$
\begin{aligned}
& \frac{\sqrt{(x+2)^{2}-8 x}}{\sqrt{x}-\frac{2}{\sqrt{x}}}=\frac{\sqrt{x^{2}+4 x+4-8 x}}{\frac{(\sqrt{x})^{2}-2}{\sqrt{x}}}=\frac{\sqrt{x} \sqrt{x^{2}-4 x-4}}{x-2}= \\
& =\frac{\sqrt{x} \sqrt{\left(x^{2}-2\right)^{2}}}{x-2}=\frac{\sqrt{x} \cdot|x-2|}{x-2}
\end{aligned}
$$

From here:

1) for $x<2, -\sqrt{x}$;
2) for $x \in(2 ;+\infty), \sqrt{x}$.

Answer: $-\sqrt{x}$ for $x \in(0 ; 2) ; \sqrt{x}$ for $x \in(2 ; \infty)$.