a: ĐKXĐ: x>=0; x<>1/4; x<>1
b: \(N=\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\left(2x+\sqrt{x}-1\right)\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\dfrac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(2\sqrt{x}-1\right)\cdot\dfrac{-1}{\left(x-\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(M=\dfrac{2\sqrt{x}-1}{\left(x-\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-\sqrt{x}+1}\)
c: Để m là số nguyên thì \(x-\sqrt{x}+1-1⋮x-\sqrt{x}+1\)
=>\(x-\sqrt{x}+1\in\left\{1;-1\right\}\)
=>\(x-\sqrt{x}+1=1\)
=>x=0(nhận) hoặc x=1(loại)