Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
M = \left( {\dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} - \dfrac{{\sqrt x - 2}}{{x - 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \left( {\dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} - \dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x - 2 - \left( {x - \sqrt x - 2} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\sqrt x .\left( {x - 1} \right)}}\\
= \dfrac{2}{{x - 1}}\\
2)\left| {M + 1} \right| > M + 1\\
\Leftrightarrow M + 1 < 0\\
\Leftrightarrow \dfrac{2}{{x - 1}} + 1 < 0\\
\Leftrightarrow \dfrac{{2 + x - 1}}{{x - 1}} < 0\\
\Leftrightarrow \dfrac{{x + 1}}{{x - 1}} < 0\\
\Leftrightarrow x - 1 < 0\\
\left( {do:x > 0 \Leftrightarrow x + 1 > 1 > 0} \right)\\
\Leftrightarrow x < 1\\
Vậy\,0 < x < 1\\
3)\sqrt M = \dfrac{1}{2}\\
\Leftrightarrow M = \dfrac{1}{4}\\
\Leftrightarrow \dfrac{2}{{x - 1}} = \dfrac{1}{4}\\
\Leftrightarrow x - 1 = 8\\
\Leftrightarrow x = 9\left( {tmdk} \right)\\
Vậy\,x = 9
\end{array}$
