a: Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{3x}{x-1}\right):\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{x+\sqrt{x}+2x-2\sqrt{x}-3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0\right\}\)
hay \(x\in\left\{4;0\right\}\)