Đáp án:
$Dkxd:x \ge 0;x \ne 1$
$\begin{array}{l}
A = \dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{2\sqrt x - 1}}{{1 - \sqrt x }} + \dfrac{{2x}}{{x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) - \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 2x}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x - 2x - \sqrt x + 1 + 2x}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}
\end{array}$
