Giải thích các bước giải:
$\begin{array}{l}
a)A = \dfrac{{x + 2\sqrt x + 1}}{{x - 1}} + \dfrac{{x - 1}}{{x - 2\sqrt x + 1}}\left( {DK:x > 0;x \ne 1} \right)\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} + \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{2\sqrt x + 2}}{{\sqrt x - 1}}\\
b)B = \dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }}\left( {DK:x > 0;x \ne 1} \right)\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\sqrt x }}\\
= 2\\
c)C = \dfrac{1}{{1 - \sqrt x }} + \dfrac{x}{{\sqrt x - 1}}\left( {DK:x \ge 0;x \ne 1} \right)\\
= \dfrac{{1 - x}}{{1 - \sqrt x }}\\
= \dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{1 - \sqrt x }}\\
= \sqrt x + 1\\
d)D = \left( {1 + \dfrac{{\sqrt x }}{{x + 1}}} \right):\dfrac{{x\sqrt x - 1}}{{\sqrt x - 1}}\left( {DK:x \ge 0;x \ne 1} \right)\\
= \left( {\dfrac{{x + \sqrt x + 1}}{{x + 1}}} \right):\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{x + \sqrt x + 1}}{{x + 1}}.\dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{1}{{x + 1}}
\end{array}$
