Đáp án:
$\begin{array}{l}
x > 0;x \ne 1\\
a)P = \frac{1}{{\sqrt x + 1}} + \frac{x}{{\sqrt x - x}}\\
= \frac{1}{{\sqrt x + 1}} + \frac{x}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \frac{1}{{\sqrt x + 1}} + \frac{{\sqrt x }}{{1 - \sqrt x }}\\
= \frac{{1 - \sqrt x + \sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {1 - \sqrt x } \right)}}\\
= \frac{{1 - \sqrt x + x + \sqrt x }}{{1 - x}}\\
= \frac{{1 + x}}{{1 - x}}\\
2)\\
x = 3\left( {tmdk} \right)\\
\Rightarrow P = \frac{{1 + 3}}{{1 - 3}} = \frac{4}{{ - 2}} = - 2\\
3)\\
x > 0;x \ne 1\\
P = \frac{{1 + x}}{{1 - x}} = \frac{{x - 1 + 2}}{{1 - x}} = - 1 + \frac{2}{{1 - x}}\\
P \in Z\\
\Rightarrow \frac{2}{{1 - x}} \in Z\\
\Rightarrow \left( {1 - x} \right) \in U\left( 2 \right) = \{ - 2; - 1;1;2\} \\
\Rightarrow x \in \{ 3;2;0; - 1\} \\
Mà:x > 0;x \ne 1\\
\Rightarrow x \in \{ 0;2;3\}
\end{array}$
