Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
A = \left( {\dfrac{{x - 2}}{{x + 2\sqrt x }} + \dfrac{1}{{\sqrt x  + 2}}} \right).\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \left( {\dfrac{{x - 2}}{{\sqrt x \left( {\sqrt x  + 2} \right)}} + \dfrac{1}{{\sqrt x  + 2}}} \right).\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {x - 2} \right) + 1.\sqrt x }}{{\sqrt x \left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{x + \sqrt x  - 2}}{{\sqrt x .\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}{{\sqrt x .\left( {\sqrt x  + 2} \right)}}.\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x }}\\
B = \left( {\dfrac{{\sqrt x  + 1}}{{x - 2\sqrt x }} - \dfrac{1}{{\sqrt x  - 2}}} \right)\left( {x - 3\sqrt x  + 2} \right)\\
 = \left( {\dfrac{{\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 2} \right)}} - \dfrac{1}{{\sqrt x  - 2}}} \right).\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) - 1.\left( {\sqrt x } \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}.\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)\\
 = \dfrac{1}{{\sqrt x \left( {\sqrt x  - 2} \right)}}.\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
C = \left( {\dfrac{1}{{\sqrt x  + 1}} - \dfrac{2}{{x\sqrt x  + x + \sqrt x  + 1}}} \right):\left( {2 - \dfrac{{2x - \sqrt x }}{{x + 1}}} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  + 1}} - \dfrac{2}{{x.\left( {\sqrt x  + 1} \right) + \left( {\sqrt x  + 1} \right)}}} \right):\dfrac{{2.\left( {x + 1} \right) - \left( {2x - \sqrt x } \right)}}{{x + 1}}\\
 = \left( {\dfrac{1}{{\sqrt x  + 1}} - \dfrac{2}{{\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}} \right):\dfrac{{2x + 2 - 2x + \sqrt x }}{{x + 1}}\\
 = \dfrac{{1.\left( {x + 1} \right) - 2}}{{\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  + 2}}{{x + 1}}\\
 = \dfrac{{x - 1}}{{\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{x + 1}}{{\sqrt x  + 2}}\\
 = \dfrac{{x - 1}}{{\left( {\sqrt x  + 1} \right).\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}
\end{array}\)