Ta có \(A=\dfrac{2\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{2\sqrt{x}+2-5}{\sqrt{x}+1}=2-\dfrac{5}{\sqrt{x}+1}\) có GT nguyên <=> \(\dfrac{5}{\sqrt{x}+1}\) có GT nguyên <=> \(5⋮\sqrt{x}+1\) <=> \(\sqrt{x}+1\in\left\{-5,-1,1,5\right\}\Leftrightarrow\sqrt{x}\in\left\{-6,-2,0,4\right\}\Leftrightarrow x\in\left\{36,4,0,16\right\}\left(tm\right)\)Vậy...