Đáp án:
\(\,\,x \in \left\{ {0;\,4;\,\,9} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
Q = \frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
DK:\,\,\,x \ge 0,\,\,x \ne 1.\\
Ta\,\,co:\,\,\,Q = \frac{{\sqrt x + 1}}{{\sqrt x - 1}} = \frac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}} = 1 + \frac{2}{{\sqrt x - 1}}\\
\Rightarrow De\,\,Q\, \in Z \Rightarrow \frac{2}{{\sqrt x - 1}} \in Z\\
\Rightarrow \sqrt x - 1 \in U\left( 2 \right)\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x - 1 = - 2\\
\sqrt x - 1 = - 1\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = - 1\,\,\left( {ktm} \right)\\
\sqrt x = 0\\
\sqrt x = 2\\
\sqrt x = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\,\,\,\left( {tm} \right)\\
x = 4\,\,\left( {tm} \right)\\
x = 9\,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,\,x \in \left\{ {0;\,4;\,\,9} \right\}.
\end{array}\)
