Đáp án:
$\begin{array}{l}
a)\dfrac{{x - 4}}{{\sqrt x - 2}} + \dfrac{{x + 2\sqrt x }}{{\sqrt x + 2}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x - 2}} + \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}\\
= \sqrt x + 2 + \sqrt x \\
= 2\sqrt x + 2\\
b)\dfrac{{x\sqrt x + 1}}{{x - \sqrt x + 1}} - \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}}\\
= \sqrt x + 1 - \left( {\sqrt x - 1} \right)\\
= 2\\
c)\left( {\dfrac{{a - 2\sqrt a }}{{\sqrt a - 2}} - 3} \right)\left( {\dfrac{{a + 3\sqrt a }}{{\sqrt a + 3}} + 3} \right)\\
= \left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)\\
= a - 9\\
d)\left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right)\left( {\dfrac{{x - 1}}{{4\sqrt x }}} \right)\\
= \dfrac{{\sqrt x - 1 - \sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{x - 1}}{{4\sqrt x }}\\
= \dfrac{{ - 2}}{{4\sqrt x }}\\
= \dfrac{{ - 1}}{{2\sqrt x }}\\
e)\dfrac{{\sqrt x }}{{\sqrt x + 2}} + \dfrac{{\sqrt x }}{{\sqrt x - 2}} - \dfrac{{2x + 4}}{{x - 4}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) + \sqrt x \left( {\sqrt x + 2} \right) - 2x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 2\sqrt x + x + 2\sqrt x - 2x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{ - 4}}{{x - 4}}\\
= \dfrac{4}{{4 - x}}\\
f)\left( {\dfrac{{\sqrt a - 4}}{{a - 2\sqrt a }} - \dfrac{3}{{2 - \sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 2}}{{\sqrt a }} - \dfrac{{\sqrt a }}{{\sqrt a - 2}}} \right)\\
= \dfrac{{\sqrt a - 4 + 3\sqrt a }}{{\sqrt a \left( {\sqrt a - 2} \right)}}:\dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right) - \sqrt a .\sqrt a }}{{\sqrt a \left( {\sqrt a - 2} \right)}}\\
= \dfrac{{4\sqrt a - 4}}{{\sqrt a \left( {\sqrt a - 2} \right)}}.\dfrac{{\sqrt a \left( {\sqrt a - 2} \right)}}{{a - 4 - a}}\\
= \dfrac{{4\left( {\sqrt a - 1} \right)}}{{ - 4}}\\
= 1 - \sqrt a
\end{array}$
