\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{x+\sqrt{x}-2}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)-3\left(\sqrt{x}-1\right)+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
Ta có: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{x+\sqrt{x}-2}\)
\(=\dfrac{x-4-3\left(\sqrt{x}-1\right)+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-1-3\sqrt{x}+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
