Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
P = \frac{{x - 3\sqrt x + 2}}{{x - 1}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
b)x = 4\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 2\\
\Rightarrow P = \frac{{2 - 2}}{{2 + 1}} = 0\\
c)P = \frac{1}{4}\\
\Rightarrow \frac{{\sqrt x - 2}}{{\sqrt x + 1}} = \frac{1}{4}\\
\Rightarrow 4\sqrt x - 8 = \sqrt x + 1\\
\Rightarrow 3\sqrt x = 9\\
\Rightarrow \sqrt x = 3\\
\Rightarrow x = 9\left( {tmdk} \right)\\
d)P = \frac{{\sqrt x - 2}}{{\sqrt x + 1}} = \frac{{\sqrt x + 1 - 3}}{{\sqrt x + 1}} = 1 - \frac{3}{{\sqrt x + 1}}\\
P \in Z\\
\Rightarrow \frac{3}{{\sqrt x + 1}} \in Z\\
\Rightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;3} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Rightarrow x \in \left\{ {0;4} \right\}\left( {tmdk} \right)
\end{array}$
