Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
a,\\
P = \left( {\dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} - \dfrac{{\sqrt x - 2}}{{x - 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \left( {\dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} - \dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right).\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {x + \sqrt x - 2} \right) - \left( {x - \sqrt x - 2} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{2}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{2}{{x - 1}}\\
b,\\
P \in Z \Leftrightarrow \dfrac{2}{{x - 1}} \in Z\\
x \in Z \Rightarrow \left( {x - 1} \right) \in \left\{ { \pm 1; \pm 2} \right\}\\
\Rightarrow x \in \left\{ { - 1;0;2;3} \right\}\\
x > 0,x \ne 1 \Rightarrow x \in \left\{ {2;3} \right\}
\end{array}\)
