Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 25\\
\left| A \right| > A\\
\Rightarrow A < 0\\
\Rightarrow \frac{{\sqrt x }}{{\sqrt x - 5}} < 0\\
\Rightarrow 0 < \sqrt x < 5\\
\Rightarrow 0 < x < 25\\
Vậy\,0 < x < 25\,thì\,\left| A \right| > A\\
b)Dkxd:x \ge 0;x \ne 25\\
\left| A \right| = - A\\
\Rightarrow A \le 0\\
\Rightarrow \frac{{x - 10\sqrt x + 25}}{{x - 25}} \le 0\\
\Rightarrow \frac{{{{\left( {\sqrt x - 5} \right)}^2}}}{{x - 25}} \le 0\\
\Rightarrow x < 25\\
Vậy\,0 \le x < 25\,thì\,\left| A \right| = - A
\end{array}$
