Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
a,\\
\dfrac{1}{P} = \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}\,\,\,\,\,\,\,\,\left( {x \ne 1} \right)\\
\Leftrightarrow \dfrac{1}{P} = \dfrac{{\left( {\sqrt x - 1} \right) + 3}}{{\sqrt x - 1}}\\
\Leftrightarrow \dfrac{1}{P} = 1 + \dfrac{3}{{\sqrt x - 1}}\\
\dfrac{1}{P} \in Z \Leftrightarrow 1 + \dfrac{3}{{\sqrt x - 1}} \in Z \Leftrightarrow \dfrac{3}{{\sqrt x - 1}} \in Z\\
\Rightarrow \sqrt x - 1 \in \left\{ { \pm 1; \pm 3} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ { - 2;\,\,0;\,\,2;\,\,4} \right\}\\
\sqrt x \ge 0 \Rightarrow \sqrt x \in \left\{ {0;2;4} \right\} \Rightarrow x \in \left\{ {0;4;16} \right\}\\
b,\\
P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} = \dfrac{{\left( {\sqrt x + 2} \right) - 3}}{{\sqrt x + 2}} = 1 - \dfrac{3}{{\sqrt x + 2}}\\
P \in Z \Leftrightarrow 1 - \dfrac{3}{{\sqrt x + 2}} \in Z \Leftrightarrow \dfrac{3}{{\sqrt x + 2}} \in Z\\
\Rightarrow \sqrt x + 2 \in \left\{ { \pm 1; \pm 3} \right\}\\
\Rightarrow \sqrt x \in \left\{ { - 5;\,\, - 3;\,\, - 1;\,\,1} \right\}\\
\sqrt x \ge 0 \Rightarrow \sqrt x = 1 \Leftrightarrow x = 1
\end{array}\)
