Bài giải:

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

(Vì 0 < x < 1 nên $\sqrt{x}-1<0\Rightarrow \left| \frac{\sqrt{x}-1}{\sqrt{x}+1} \right|=-\frac{\sqrt{x}-1}{\sqrt{x}+1}$ )