Đáp án:
$\begin{array}{l}
a)Dlxd:x > 0;x \ne 1\\
A = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{{x + \sqrt x }}{{x - 1}}} \right).\left( {\dfrac{1}{{\sqrt x }} - 1} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) - x - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{1 - \sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - \sqrt x - x - \sqrt x }}{{\sqrt x + 1}}.\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}}.\dfrac{{ - 1}}{{\sqrt x }}\\
= \dfrac{2}{{\sqrt x + 1}}\\
b)x = \dfrac{2}{{2 + \sqrt 3 }}\\
= \dfrac{{2\left( {2 - \sqrt 3 } \right)}}{{{2^2} - 3}}\\
= 4 - 2\sqrt 3 \\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 - 1\\
A = \dfrac{2}{{\sqrt x + 1}} = \dfrac{2}{{\sqrt 3 - 1 + 1}} = \dfrac{2}{{\sqrt 3 }} = \dfrac{{2\sqrt 3 }}{3}
\end{array}$
