Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 25\\
A = \dfrac{{\sqrt x }}{{\sqrt x - 5}} - \dfrac{{10\sqrt x }}{{x - 25}} - \dfrac{5}{{\sqrt x + 5}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 5} \right) - 10\sqrt x - 5\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{x + 5\sqrt x - 10\sqrt x - 5\sqrt x + 25}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{x - 10\sqrt x + 25}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 5} \right)}^2}}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}}\\
b)x = 9\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 3\\
\Rightarrow A = \dfrac{{3 - 5}}{{3 + 5}} = \dfrac{{ - 1}}{4}\\
c)A < \dfrac{1}{3}\\
\Rightarrow \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}} < \dfrac{1}{3}\\
\Rightarrow \dfrac{{3\sqrt x - 15 - \sqrt x - 5}}{{3\left( {\sqrt x + 5} \right)}} < 0\\
\Rightarrow \dfrac{{2\sqrt x - 20}}{{3\left( {\sqrt x + 5} \right)}} < 0\\
\Rightarrow 2\sqrt x - 20 < 0\\
\Rightarrow \sqrt x < 10\\
\Rightarrow x < 100\\
Vậy\,0 \le x < 100;x \ne 25
\end{array}$
