Có \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-4}{\sqrt{x}+2}=2-\dfrac{4}{\sqrt{x}+2}\)
Để A nguyên => \(\sqrt{x}+2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2\in\left\{2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\)
\(\Leftrightarrow x\in\left\{0;4\right\}\)
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