Đáp án:
$\begin{array}{l}
a)A = \frac{{6 - \sqrt x }}{{x - 9\sqrt x + 20}} + \frac{2}{{\sqrt x - 4}} - \frac{3}{{\sqrt x - 5}}\\
= \frac{{6 - \sqrt x + 2\left( {\sqrt x - 5} \right) - 3\left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x - 5} \right)}}\\
= \frac{{6 - \sqrt x + 2\sqrt x - 10 - 3\sqrt x + 12}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x - 5} \right)}}\\
= \frac{{8 - 2\sqrt x }}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x - 5} \right)}}\\
= \frac{{ - 2\left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x - 5} \right)}}\\
= \frac{{ - 2}}{{\sqrt x - 5}}\\
b)x = 121\left( {tmdk} \right)\\
\Rightarrow x = {11^2}\\
\Rightarrow \sqrt x = 11\\
\Rightarrow A = \frac{{ - 2}}{{11 - 5}} = - \frac{1}{3}\\
c)Dkxd:x \ge 0;x \ne 16;x \ne 25\\
A \ge \frac{1}{2}\\
\Rightarrow \frac{{ - 2}}{{\sqrt x - 5}} \ge \frac{1}{2}\\
\Rightarrow \frac{{ - 2.2 - \sqrt x + 5}}{{2\left( {\sqrt x - 5} \right)}} \ge 0\\
\Rightarrow \frac{{1 - \sqrt x }}{{\sqrt x - 5}} \ge 0\\
\Rightarrow 1 \le \sqrt x < 5\\
\Rightarrow 1 \le x < 25\\
Vậy\,1 \le x < 25;x \ne 16
\end{array}$
