a: \(A=\dfrac{x-2\sqrt{x}}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\dfrac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(x-2\sqrt{x}\right)+\left(\sqrt{x}-1\right)+2x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}-1+2x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\cdot\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\cdot\sqrt{x}}=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: \(A\in Z\)
=>\(\sqrt{x}+1⋮x+\sqrt{x}+1\)
=>\(\sqrt{x}\left(\sqrt{x}+1\right)+1-1⋮x+\sqrt{x}+1\)
=>\(-1⋮x+\sqrt{x}+1\)
=>\(x+\sqrt{x}+1\in\left\{1;-1\right\}\)
=>\(x+\sqrt{x}+1=1\)
=>\(x+\sqrt{x}=0\)
=>x=0(loại)
