Giải thích các bước giải:
ĐKXĐ: $x>0;x\ne 1$
Ta có:
$\begin{array}{l}
M = \left( {\dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \left( {\dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right){{\left( {\sqrt x + 1} \right)}^2}}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{ - x\sqrt x + x + 4\sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\sqrt x \left( { - x + \sqrt x + 4} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{ - x + \sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - x + \sqrt x + 4}}{{x - 1}}
\end{array}$
Vậy $M = \dfrac{{ - x + \sqrt x + 4}}{{x - 1}}$ với $x>0;x\ne 1$
