Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
P = \left( {\dfrac{1}{{x + \sqrt x }} - \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x }}{{1 + x + 2\sqrt x }}\\
 = \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x  + 1} \right)}} - \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x }}{{x + 2\sqrt x  + 1}}\\
 = \dfrac{{1 - \sqrt x }}{{\sqrt x .\left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x }}{{{{\left( {\sqrt x  + 1} \right)}^2}}}\\
 = \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}\\
 = \dfrac{{\left( {1 - \sqrt x } \right).\left( {\sqrt x  + 1} \right)}}{{{{\sqrt x }^2}}}\\
 = \dfrac{{1 - x}}{x}\\
Q = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 1}} - \dfrac{{\sqrt x }}{{x + \sqrt x }}} \right):\dfrac{{\sqrt x  - 1}}{{x - 1}}\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 1}} - \dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}} \right):\dfrac{{\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 1}} - \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}.\left( {\sqrt x  + 1} \right)\\
 = \sqrt x  - 1\\
N = \left( {\dfrac{1}{{\sqrt x  - 3}} + \dfrac{1}{{\sqrt x  + 3}}} \right)\left( {1 - \dfrac{3}{{\sqrt x }}} \right)\\
 = \dfrac{{1.\left( {\sqrt x  + 3} \right) + 1.\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x }}\\
 = \dfrac{{2\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x }}\\
 = \dfrac{2}{{\sqrt x  + 3}}\\
M = \dfrac{{9\sqrt x  - \sqrt {25x}  + \sqrt {4{x^3}} }}{{{x^2} + 2x}}\\
 = \dfrac{{9\sqrt x  - 5\sqrt x  + 2x\sqrt x }}{{{x^2} + 2x}}\\
 = \dfrac{{2x\sqrt x  + 4\sqrt x }}{{x\left( {x + 2} \right)}}\\
 = \dfrac{{2\sqrt x \left( {x + 2} \right)}}{{x.\left( {x + 2} \right)}}\\
 = \dfrac{{2\sqrt x }}{x} = \dfrac{2}{{\sqrt x }}
\end{array}\)