Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 1\\
P = \left( {1 + \dfrac{{\sqrt x }}{{x + 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x\sqrt x + \sqrt x - x - 1}}} \right) - 1\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\dfrac{{x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}} - 1\\
= \dfrac{{x + \sqrt x + 1}}{{x + 1}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} - 1\\
= \left( {x + \sqrt x + 1} \right).\dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - 1\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}} - 1\\
= \dfrac{{x + \sqrt x + 1 - \sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{x}{{\sqrt x + 1}}\\
b)Q = \sqrt x - P\\
= \sqrt x - \dfrac{x}{{\sqrt x + 1}}\\
= \dfrac{{x + \sqrt x - x}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1 - 1}}{{\sqrt x + 1}}\\
= 1 - \dfrac{1}{{\sqrt x + 1}}\\
Q \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 = 1\left( {do:\sqrt x + 1 \ge 1} \right)\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0
\end{array}$
