Để $A$ nguyên thì
$\begin{array}{l} A = \dfrac{{4\sqrt x + 16}}{{\sqrt x + 2}} = \dfrac{{4\left( {\sqrt x + 2} \right) + 8}}{{\sqrt x + 2}}\\ A = 4 + \dfrac{8}{{\sqrt x + 2}} \in Z\\ \Rightarrow 8 \vdots \sqrt x + 2\\ \Rightarrow \sqrt x + 2 \in U\left( 8 \right) = \left\{ {2;4;8} \right\}\left( {do\,\sqrt x + 2 \ge 2} \right)\\ \Rightarrow x \in \left\{ {0;4;36} \right\} \end{array}$