Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
b)P = \left( {\frac{{x - 2}}{{x + 2\sqrt x }} + \frac{1}{{\sqrt x + 2}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \left( {\frac{{x - 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}} + \frac{1}{{\sqrt x + 2}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \frac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \frac{{\sqrt x + 1}}{{\sqrt x }}\\
c)2P = 2\sqrt x + 5\\
\Rightarrow 2.\frac{{\sqrt x + 1}}{{\sqrt x }} = 2\sqrt x + 5\\
\Rightarrow 2\sqrt x + 2 = \sqrt x \left( {2\sqrt x + 5} \right)\\
\Rightarrow 2x + 3\sqrt x - 2 = 0\\
\Rightarrow 2x + 4\sqrt x - \sqrt x - 2 = 0\\
\Rightarrow \left( {\sqrt x + 2} \right)\left( {2\sqrt x - 1} \right) = 0\\
\Rightarrow \sqrt x = \frac{1}{2}\left( {do:\sqrt x > 0} \right)\\
\Rightarrow x = \frac{1}{4}\left( {tmdk} \right)
\end{array}$
