Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
P = \dfrac{{{x^2} - \sqrt x }}{{x + \sqrt x + 1}} - \dfrac{{2x + \sqrt x }}{{\sqrt x }} + \dfrac{{2\left( {x - 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {x\sqrt x - 1} \right)}}{{x + \sqrt x + 1}} - \dfrac{{\sqrt x \left( {2\sqrt x + 1} \right)}}{{\sqrt x }} + \dfrac{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}} - \left( {2\sqrt x + 1} \right) + 2\left( {\sqrt x + 1} \right)\\
= \sqrt x \left( {\sqrt x - 1} \right) - 2\sqrt x - 1 + 2\sqrt x + 2\\
= x - \sqrt x + 1\\
b)Q = \dfrac{{2\sqrt x }}{P} = \dfrac{{2\sqrt x }}{{x - \sqrt x + 1}} > 0\\
\Leftrightarrow Q.x - Q.\sqrt x + Q = 2\sqrt x \\
\Leftrightarrow Q.x - \left( {Q + 2} \right).\sqrt x + Q = 0\\
\Leftrightarrow \Delta ' \ge 0\\
\Leftrightarrow {\left( {Q + 2} \right)^2} - 4{Q^2} \ge 0\\
\Leftrightarrow \left( {Q + 2 - 2Q} \right)\left( {Q + 2 + 2Q} \right) \ge 0\\
\Leftrightarrow \left( {Q - 2} \right)\left( {3Q + 2} \right) \le 0\\
\Leftrightarrow 0 < Q \le 2\\
Khi:Q \in Z\\
\Leftrightarrow \left[ \begin{array}{l}
Q = 1\\
Q = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3\sqrt x + 1 = 0\\
2x - 4\sqrt x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = \dfrac{{3 \pm \sqrt 5 }}{2}\\
x - 2\sqrt x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7 \pm 3\sqrt 5 }}{2}\left( {tm} \right)\\
\sqrt x = 1\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{{7 \pm 3\sqrt 5 }}{2}
\end{array}$
