Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
M = \left( {\dfrac{1}{{\sqrt x  - 1}} - \dfrac{1}{{\sqrt x }}} \right)\left( {\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 2}} - \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}}} \right)\left( {DK:x \ge 0;x \ne \left\{ {0;1;4} \right\}} \right)\\
 = \dfrac{{\sqrt x  - \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\sqrt x }}.\dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{1}{{\sqrt x \left( {\sqrt x  - 1} \right)}}.\dfrac{3}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{3}{{\sqrt x \left( {\sqrt x  - 2} \right){{\left( {\sqrt x  - 1} \right)}^2}}}\\
 = \dfrac{3}{{\left( {x - 2\sqrt x } \right)\left( {x - 2\sqrt x  + 1} \right)}}
\end{array}$

Vậy $M = \dfrac{3}{{\left( {x - 2\sqrt x } \right)\left( {x - 2\sqrt x  + 1} \right)}}$ với ${x \ge 0;x \ne \left\{ {0;1;4} \right\}}$

b) Ta có:

$\begin{array}{l}
M = \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{3}{{\left( {x - 2\sqrt x } \right)\left( {x - 2\sqrt x  + 1} \right)}} = \dfrac{1}{3}\\
 \Leftrightarrow \left( {x - 2\sqrt x } \right)\left( {x - 2\sqrt x  + 1} \right) - 9 = 0\\
 \Leftrightarrow {\left( {x - 2\sqrt x } \right)^2} + \left( {x - 2\sqrt x } \right) - 9 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2\sqrt x  = \dfrac{{ - 1 + \sqrt {37} }}{2}\left( {c;Do:x - 2\sqrt x  \ge  - 1} \right)\\
x - 2\sqrt x  = \dfrac{{ - 1 - \sqrt {37} }}{2}\left( {l;Do:x - 2\sqrt x  \ge  - 1} \right)
\end{array} \right.\\
 \Leftrightarrow x - 2\sqrt x  = \dfrac{{ - 1 + \sqrt {37} }}{2}\\
 \Leftrightarrow {\left( {\sqrt x } \right)^2} - 2\sqrt x  + \dfrac{{1 - \sqrt {37} }}{2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 1 + \sqrt {\dfrac{{1 + \sqrt {37} }}{2}} \left( c \right)\\
\sqrt x  = 1 - \sqrt {\dfrac{{1 + \sqrt {37} }}{2}} \left( l \right)
\end{array} \right.\\
 \Leftrightarrow \sqrt x  = 1 + \sqrt {\dfrac{{1 + \sqrt {37} }}{2}} \\
 \Leftrightarrow x = {\left( {1 + \sqrt {\dfrac{{1 + \sqrt {37} }}{2}} } \right)^2} (tm)
\end{array}$

Vậy $x = {\left( {1 + \sqrt {\dfrac{{1 + \sqrt {37} }}{2}} } \right)^2}$