Đáp án:

$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
P = \dfrac{{{x^2} + 2}}{{1 - {x^3}}} - \dfrac{1}{{2\left( {1 + \sqrt x } \right)}} - \dfrac{1}{{2\left( {1 - \sqrt x } \right)}}\\
 = \dfrac{{{x^2} + 2}}{{1 - {x^3}}} - \dfrac{{1 - \sqrt x  + 1 + \sqrt x }}{{2\left( {1 + \sqrt x } \right)\left( {1 - \sqrt x } \right)}}\\
 = \dfrac{{{x^2} + 2}}{{1 - {x^3}}} - \dfrac{2}{{2\left( {1 - x} \right)}}\\
 = \dfrac{{{x^2} + 2}}{{\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}} - \dfrac{1}{{1 - x}}\\
 = \dfrac{{{x^2} + 2 - \left( {1 + x + {x^2}} \right)}}{{\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}\\
 = \dfrac{{{x^2} + 2 - 1 - x - {x^2}}}{{\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}\\
 = \dfrac{{1 - x}}{{\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}\\
 = \dfrac{1}{{1 + x + {x^2}}}\\
b)Q = \dfrac{1}{{\left( {x - 1} \right).P}} = \dfrac{1}{{\left( {x - 1} \right).\dfrac{1}{{1 + x + {x^2}}}}}\\
 = \dfrac{{1 + x + {x^2}}}{{x - 1}}\\
 = \dfrac{{{x^2} - x + 2x - 2 + 3}}{{x - 1}}\\
 = x + 2 + \dfrac{3}{{x - 1}}\\
Q \in Z\\
 \Rightarrow \dfrac{3}{{x - 1}} \in Z\\
 \Rightarrow \left( {x - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow x \in \left\{ { - 2;0;2;4} \right\}\\
Do:x \ge 0;x \ne 1\\
 \Rightarrow x \in \left\{ {0;2;4} \right\}\\
\text{Vậy}\,x \in \left\{ {0;2;4} \right\}
\end{array}$