ĐKXĐ: \(x\ge0;x\ne\left\{1;\frac{1}{4}\right\}\)
\(M=1-\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\left(\frac{-\sqrt{x}\left(\sqrt{x}-1\right)^2}{2\sqrt{x}-1}\right)\)
\(=1-\left(\frac{-1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-\sqrt{x}+1}\right)\left(\frac{-\sqrt{x}\left(\sqrt{x}-1\right)^2\left(2\sqrt{x}-1\right)}{2\sqrt{x}-1}\right)\)
\(=1-\left(\frac{-x+\sqrt{x}-1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\left(-\sqrt{x}\left(\sqrt{x}-1\right)^2\right)\)
\(=1-\left(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\sqrt{x}\left(\sqrt{x}-1\right)^2\)
\(=1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)^3}{x-\sqrt{x}+1}=1-\frac{x^2-3x\sqrt{x}+3x-\sqrt{x}}{x-\sqrt{x}+1}=\frac{-x^2+3x\sqrt{x}-2x+1}{x-\sqrt{x}+1}\)
Bạn coi lại đề, ko rút gọn được :(