$2.330 A=\frac{(x+2) \sqrt{(x+2)^{2}-8 x}}{x^{2}-4|x-1|}$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

$2.330 A=\frac{(x+2) \sqrt{(x+2)^{2}-8 x}}{x^{2}-4|x-1|}$.
Solution. $A=\frac{(x+2) \sqrt{x^{2}+4 x+4-8 x}}{x^{2}-4|x-1|}=\frac{(x+2) \cdot \sqrt{(x-2)^{2}}}{x^{2}-4 \cdot|x-1|}=\frac{(x+2)|x-2|}{x^{2}-4|x-1|}$.

Consider 3 cases.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-020.jpg?height=155&width=572&top_left_y=254&top_left_x=113)

2) $\left\{\begin{array}{l}12, \\ A=\frac{(x+2)(x-2)}{x^{2}-4 x+4}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>2, \\ A=\frac{(x+2)(x-2)}{(x-2)^{2}}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>2, \\ A=\frac{x+2}{x-2} .\end{array}\right.\right.\right.$

Answer: if $x \leq 1, x \neq-2 \pm 2 \sqrt{2}$, then $A=\frac{4-x^{2}}{x^{2}+4 x-4}$; if $12$, then $A=\frac{x+2}{x-2}$.