a)
\(N=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)
b)
\(N=x-\sqrt{x}+1=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy Min N = \(\frac{3}{4}\)khi x=\(\frac{1}{4}\)
Câu c) mk ko bt, sorry nha :<
\(M=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{2\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}>0\) \(\forall x>0\)
\(M-2=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{-2\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\frac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}< 0\) \(\forall x\ne1;x>0\)
\(\Rightarrow0< M< 2\Rightarrow\) để M nguyên thì \(M=1\)
\(\Rightarrow\frac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow x-3\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\) \(\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)