\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}+2-2-1}{\sqrt{x}+1}=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}\)
(đkxđ: \(x\ge0\))
Để \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+1}\in Z\)
`=>` \(\sqrt{x}+1\inƯ_{\left(3\right)}=\left\{-3;-1;1;3\right\}\)
`=>` \(\sqrt{x}\in\left\{-4;-2;0;2\right\}\)
`=>` \(x\in\left\{16;4;0\right\}\)