Đáp án:$\,x \in \left\{ {0;1;4;25} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge 0\\
A = \dfrac{{2\sqrt x + 8}}{{\sqrt x + 1}} = \dfrac{{2\sqrt x + 2 + 6}}{{\sqrt x + 1}}\\
= \dfrac{{2.\left( {\sqrt x + 1} \right) + 6}}{{\sqrt x + 1}}\\
= 2 + \dfrac{6}{{\sqrt x + 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{6}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;2;3;6} \right\}\left( {do:\sqrt x + 1 \ge 1} \right)\\
\Leftrightarrow \sqrt x \in \left\{ {0;1;2;5} \right\}\\
\Leftrightarrow x \in \left\{ {0;1;4;25} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {0;1;4;25} \right\}
\end{array}$
