Đáp án:
$\begin{array}{l}
a)A = \left( {\frac{1}{{\sqrt x + 2}} + \frac{7}{{x - 4}}} \right):\frac{1}{{\sqrt x - 2}}\\
= \frac{{\sqrt x - 2 + 7}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\left( {\sqrt x - 2} \right)\\
= \frac{{\sqrt x + 5}}{{\sqrt x + 2}}\\
b)x = \sqrt {\frac{2}{{2 - \sqrt 3 }}} - \sqrt {\frac{2}{{2 + \sqrt 3 }}} \left( {tmdk} \right)\\
= \sqrt {\frac{{2\left( {2 + \sqrt 3 } \right)}}{{4 - 3}}} - \sqrt {\frac{{2\left( {2 - \sqrt 3 } \right)}}{{4 - 3}}} \\
= \sqrt {4 + 2\sqrt 3 } - \sqrt {4 - 2\sqrt 3 } \\
= \sqrt 3 + 1 - \left( {\sqrt 3 - 1} \right)\\
= 2\\
\Rightarrow A = \frac{{\sqrt x + 5}}{{\sqrt x + 2}} = \frac{{\sqrt 2 + 5}}{{\sqrt 2 + 2}}\\
= \frac{{\left( {5 + \sqrt 2 } \right)\left( {2 - \sqrt 2 } \right)}}{{4 - 2}}\\
= \frac{{8 - 3\sqrt 2 }}{2}
\end{array}$
