$A=\sqrt{\frac{(x^{2}-3)^{2}+12x^{2}}{x^{2}}}+\sqrt{(x+2)^{2}-8x}$
$\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}=\sqrt{x^2+6+\frac{9}{x^2}}=\sqrt{(x-2)^2}+\sqrt{(x+\frac{3}{x})^2}=\left | x-2 \right |+x+\frac{3}{x}$ (chú ý đk)
$\rightarrow A \in Z \leftrightarrow \frac{3}{x} \in Z.$
Ok r nhé!