$2.347 A=\sqrt{x\left(x^{-1}+4 x-4\right)^{-1}}-\frac{2 x^{2}}{|2 x-1|}$, where $x>0$.
Solution. Let's denote by $B$ and transform the expression

$$
B=\sqrt{x\left(x^{-1}+4 x-4\right)^{-1}}=\sqrt{\frac{x}{\frac{1}{x}+4 x-4}}=\sqrt{\frac{x^{2}}{4 x^{2}-4 x+1}}=\frac{|x|}{|2 x-1|}=\frac{x}{|2 x-1|} .
$$

Therefore, $A=B-\frac{2 x^{2}}{|2 x-1|}=\frac{x-2 x^{2}}{|2 x-1|}=\frac{x(1-2 x)}{|1-2 x|}$.

1) $\left\{\begin{array}{l}A=x, \\ 1-2 x>0 \\ x>0\end{array}, \Leftrightarrow\left\{\begin{array}{l}A=x, \\ 0<x<\frac{1}{2}\end{array}\right.\right.$

2) $\left\{\begin{array}{l}A=-x, \\ 1-2 x<0 \\ x>\frac{1}{2}\end{array}\right.\right.$

Answer: if $0<x<\frac{1}{2}$, then $A=x$; if $x>\frac{1}{2}$, then $A=-x$.