Đáp án: 1;9;25
Giải thích các bước giải:
$\begin{array}{l}
Đkxđ:x \ge 0;x \ne 4\\
A = \frac{{\sqrt x + 1}}{{\sqrt x - 2}}\\
= \frac{{\sqrt x - 2 + 3}}{{\sqrt x - 2}}\\
= \frac{{\sqrt x - 2}}{{\sqrt x - 2}} + \frac{3}{{\sqrt x - 2}}\\
= 1 + \frac{3}{{\sqrt x - 2}}\\
A \in Z \Rightarrow \frac{3}{{\sqrt x - 2}} \in Z\\
\Rightarrow 3 \vdots \left( {\sqrt x - 2} \right)\\
\Rightarrow \sqrt x - 2 \in Ư\left( 3 \right)\\
\Rightarrow \sqrt x - 2 \in \{ - 3; - 1;1;3\} \\
\Rightarrow \sqrt x \in \{ - 1;1;3;5\} \\
Mà\,\sqrt x \ge 0 \Rightarrow \sqrt x \in \{ 1;3;5\} \\
\Rightarrow x \in \{ 1;9;25\} \left( {tmdk} \right)\\
Vậy\,x \in \{ 1;9;25\}
\end{array}$
