Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
M = \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right).\dfrac{{x - \sqrt x }}{{2\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1 + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{2\sqrt x + 1}}\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}.\dfrac{{\sqrt x }}{{2\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b)M = \dfrac{1}{3}\\
\Rightarrow \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{1}{3}\\
\Rightarrow 3\sqrt x = \sqrt x + 1\\
\Rightarrow 2\sqrt x = 1\\
\Rightarrow \sqrt x = \dfrac{1}{2}\\
\Rightarrow x = \dfrac{1}{4}\left( {tmdk} \right)
\end{array}$
