Đáp án:
$\begin{array}{l}
A = \dfrac{{2x}}{{\sqrt {x - 4} }}\\
= \dfrac{{2\left( {x - 4} \right) + 8}}{{\sqrt {x - 4} }}\\
= 2\sqrt {x - 4} + \dfrac{8}{{\sqrt {x - 4} }}\\
A \in Z\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 4} \in Z\\
\dfrac{8}{{\sqrt {x - 4} }} \in Z
\end{array} \right.\\
\Rightarrow \sqrt {x - 4} \in \left\{ {1;2;4;8} \right\}\\
\Rightarrow x - 4 \in \left\{ {1;4;16;64} \right\}\\
\Rightarrow x \in \left\{ {5;8;20;68} \right\}\\
Vậy\,x \in \left\{ {5;8;20;68} \right\}
\end{array}$
