Đáp án:
$\begin{array}{l}
2A\\
a)Dkxd:x \ge 0;x \ne 1\\
M = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{2}{x} - \dfrac{{2 - x}}{{x\sqrt x + x}}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\left( {\dfrac{2}{x} - \dfrac{{2 - x}}{{x\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{x\left( {\sqrt x + 1} \right)}}{{2\left( {\sqrt x + 1} \right) - 2 + x}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x - 1}}.\dfrac{x}{{2\sqrt x + 2 - 2 + x}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x - 1}}.\dfrac{x}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{x}{{\sqrt x - 1}}\\
b)M = - \dfrac{1}{2}\\
\Rightarrow \dfrac{x}{{\sqrt x - 1}} = - \dfrac{1}{2}\\
\Rightarrow 1 - \sqrt x = 2x\\
\Rightarrow 2x + \sqrt x - 1 = 0\\
\Rightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) = 0\\
\Rightarrow \sqrt x = \dfrac{1}{2}\\
\Rightarrow x = \dfrac{1}{4}\left( {tmdk} \right)\\
2B\\
a)Dkxd:x \ge 0\\
N = \left( {\dfrac{{x + 2}}{{x\sqrt x + 1}} - \dfrac{1}{{\sqrt x + 1}}} \right).\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{x + 2 - \left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{x + 2 - x + \sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}}\\
b)N = \dfrac{8}{9}\\
\Rightarrow \dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}} = \dfrac{8}{9}\\
\Rightarrow 3\sqrt x = 2\left( {x - \sqrt x + 1} \right)\\
\Rightarrow 2x - 5\sqrt x + 2 = 0\\
\Rightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{1}{2}\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\left( {tmdk} \right)\\
x = 4\left( {tmdk} \right)
\end{array} \right.
\end{array}$
