1: \(M=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
2: \(x=\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{2}=\dfrac{2}{2}=1\)
Khi x=1 thì \(A=\dfrac{1-1}{1+2}=0\)
\(1,M=\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\) với x ≥ 0
\(2,x=\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\\ =\dfrac{\sqrt{3}+1-\sqrt{3}+1}{3-1}=\dfrac{2}{2}=1\)
\(Th\text{ạ}yx=1v\text{ạ}oN\\ N=\dfrac{\sqrt{1}-1}{\sqrt{1}+2}=\dfrac{1-1}{1+2}=0\\ V\text{ạ}ỵ......\)
