Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
P = \left( {\dfrac{{x + 3}}{{x - 1}} + \dfrac{{\sqrt x }}{{1 - \sqrt x }} + \dfrac{1}{{1 + \sqrt x }}} \right)\left( {\dfrac{{x\sqrt x - 1}}{{\sqrt x - 1}} + \sqrt x } \right)\\
= \left( {\dfrac{{x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}}} \right)\left( {\dfrac{{{{\sqrt x }^3} - {1^3}}}{{\sqrt x - 1}} + \sqrt x } \right)\\
= \left( {\dfrac{{x + 3 - \sqrt x \left( {\sqrt x + 1} \right) + \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\left( {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x - 1}} + \sqrt x } \right)\\
= \left( {\dfrac{{x + 3 - x - \sqrt x + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\left( {x + \sqrt x + 1 + \sqrt x } \right)\\
= \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.{\left( {\sqrt x + 1} \right)^2}\\
= \dfrac{{2.\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
2,\\
\dfrac{1}{2}P = \dfrac{1}{2}.\dfrac{{2\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}} = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\left( {\sqrt x - 1} \right) + 2}}{{\sqrt x - 1}} = 1 + \dfrac{2}{{\sqrt x - 1}}\\
\dfrac{1}{2}P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
x \in Z \Rightarrow \left( {\sqrt x - 1} \right) \in \left\{ { \pm 1; \pm 2} \right\}\\
\Rightarrow \sqrt x \in \left\{ { - 1;0;2;3} \right\}\\
\sqrt x \ge 0 \Rightarrow \sqrt x \in \left\{ {0;2;3} \right\} \Rightarrow x \in \left\{ {0;4;9} \right\}\,\,\,\,\,\left( {t/m} \right)
\end{array}\)
