Đáp án:
$\begin{array}{l}
A = \left( {\frac{{x + 2\sqrt x }}{{x - 2\sqrt x }} + \frac{{\sqrt x }}{{\sqrt x - 2}}} \right).\frac{1}{{\sqrt x + 1}}\\
a)dkxd:\left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
A = \left( {\frac{{x + 2\sqrt x }}{{x - 2\sqrt x }} + \frac{{\sqrt x }}{{\sqrt x - 2}}} \right).\frac{1}{{\sqrt x + 1}}\\
= \left( {\frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}} + \frac{{\sqrt x }}{{\sqrt x - 2}}} \right).\frac{1}{{\sqrt x + 1}}\\
= \frac{{\sqrt x + 2 + \sqrt x }}{{\sqrt x - 2}}.\frac{1}{{\sqrt x + 1}}\\
= \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x - 2}}.\frac{1}{{\sqrt x + 1}}\\
= \frac{{\sqrt x }}{{\sqrt x - 2}}\\
b)x > 0;x \ne 4\\
A < 0\\
\Rightarrow \frac{{\sqrt x }}{{\sqrt x - 2}} < 0\\
\Rightarrow \sqrt x - 2 < 0\left( {do:\sqrt x > 0} \right)\\
\Rightarrow \sqrt x < 2\\
\Rightarrow x < 4\\
Vậy\,0 < x < 4
\end{array}$
