Đặt \(\sqrt{x}=a\)
Khi đó M=\(\frac{a^2}{a-1}+\frac{2a^2-a}{a-a^2}\)( ĐKXĐ : a \(\ne1,a\ne0\))
=> M = \(\frac{a^2}{a-1}+\frac{a\left(2a-1\right)}{a\left(1-a\right)}\)= \(\frac{a^2}{a-1}+\frac{2a-1}{1-a}=\frac{a^2}{a-1}+\frac{1-2a}{a-1}\)
= \(\frac{a^2-2a+1}{a-1}\)=\(\frac{\left(a-1\right)^2}{a-1}=a-1\)= \(\sqrt{x}-1\)
Vậy M = \(\sqrt{x}-1\)
M = \(\frac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\) + \(\frac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
M = \(2\sqrt{x}-1\)