Đáp án:
x=4
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{x}{{\sqrt x - 1}} = \dfrac{{x - 1 + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1}}{{\sqrt x - 1}}\\
= \sqrt x + 1 + \dfrac{1}{{\sqrt x - 1}}\\
\dfrac{x}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \sqrt x - 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
