Đáp án:
$\begin{array}{l}
A = \left( {\frac{1}{{1 - \sqrt x }} - \frac{1}{{1 + \sqrt x }}} \right).\left( {1 - \frac{1}{{\sqrt x }}} \right)\left( {x > 0;x \ne 1} \right)\\
= \frac{{1 + \sqrt x - \left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}.\frac{{\sqrt x - 1}}{{\sqrt x }}\\
= \frac{{2\sqrt x }}{{ - \left( {1 + \sqrt x } \right)}}.\frac{1}{{\sqrt x }}\\
= \frac{{ - 2}}{{1 + \sqrt x }}\\
B = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}.\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}} - \frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right)\left( {x > 0;x \ne 1} \right)\\
= {\left( {\frac{{\sqrt x .\sqrt x - 1}}{{2\sqrt x }}} \right)^2}.\frac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{{{\left( {\sqrt x - 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{{4x}}.\frac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 1} \right)}}{{4x}}.\frac{{4\sqrt x }}{1}\\
= \frac{{x - 1}}{{\sqrt x }}\\
C = \left( {\frac{1}{{\sqrt a - 3}} + \frac{1}{{\sqrt a + 3}}} \right).\left( {1 - \frac{3}{{\sqrt a }}} \right)\left( {a > 0;a \ne 9} \right)\\
= \frac{{\sqrt a + 3 + \sqrt a - 3}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)}}.\frac{{\sqrt a - 3}}{{\sqrt a }}\\
= \frac{{2\sqrt a }}{{\sqrt a \left( {\sqrt a + 3} \right)}}\\
= \frac{2}{{\sqrt a + 3}}\\
C > \frac{1}{2}\\
\Rightarrow \frac{2}{{\sqrt a + 3}} > \frac{1}{2}\\
\Rightarrow \sqrt a + 3 > 4\\
\Rightarrow \sqrt a > 1\\
\Rightarrow a > 1\\
Vậy\,a > 1;a \ne 9
\end{array}$
