Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{3 - \sqrt 5 }}{2}\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{2}{{x + \sqrt x + 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{x + \sqrt x + 1}} \in Z\\
\Leftrightarrow x + \sqrt x + 1 \in U\left( 2 \right)\\
Ma:x + \sqrt x + 1 > 0\forall x \ge 0\\
\to \left[ \begin{array}{l}
x + \sqrt x + 1 = 2\\
x + \sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + \sqrt x - 1 = 0\\
x + \sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{ - 1 + \sqrt 5 }}{2}\\
\sqrt x = \dfrac{{ - 1 - \sqrt 5 }}{2}\left( l \right)\\
\sqrt x \left( {\sqrt x + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{ - 1 + \sqrt 5 }}{2}\\
\sqrt x = 0\\
\sqrt x = - 1\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{3 - \sqrt 5 }}{2}\\
x = 0
\end{array} \right.
\end{array}\)
