ĐKXĐ : \(x>0;x\ne1\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\sqrt{x}-1\)
a) Ta có: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1\)
`A=x/(\sqrt{x}-1)-(2x-\sqrt{x})/(x-\sqrt{x})`
`A=x/(\sqrt{x}-1)-(\sqrt{x}(2\sqrt{x}-1))/(\sqrt{x}(\sqrt{x}-1))`
`A=x/(\sqrt{x}-1)-(2\sqrt{x}-1)/(\sqrt{x}-1)`
`A=(x-2\sqrt{x}+1)/(\sqrt{x}-1)`
`A=(\sqrt{x}-1)^2/(\sqrt{x}-1)`
`A=\sqrt{x}-1`
