Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 1\\
P = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{x\sqrt x - x + \sqrt x - 1}}} \right)\\
:\left( {\dfrac{{x + \sqrt x }}{{x\sqrt x + x + \sqrt x + 1}} + \dfrac{1}{{x + 1}}} \right)\\
= \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}} \right)\\
:\left( {\dfrac{{x + \sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{1}{{x + 1}}} \right)\\
= \dfrac{{x + 1 - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}:\dfrac{{x + \sqrt x + \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}}{{x + 2\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1}}{{x + 1}}.\dfrac{{x + 1}}{{\sqrt x + 1}}\\
= 1\\
b)P = \sqrt x - 2\\
\Leftrightarrow 1 = \sqrt x - 2\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow x = 9\left( {tmdk} \right)
\end{array}$
Vậy $x = 9$
