\(\sqrt{\frac{8-4\sqrt{3}}{\sqrt{6}-\sqrt{2}}}\cdot\sqrt{\sqrt{6}+\sqrt{2}}=2\)  \(=\) \(2\) nha bạn . 

 = \(\sqrt{\frac{\left(\sqrt{6}-\sqrt{2}\right)^2}{\sqrt{6}-\sqrt{2}}}\)     .      \(\sqrt{\sqrt{6} +\sqrt{2}}\) = \(\sqrt{\sqrt{6}-\sqrt{2}}\) .    \(\sqrt{\sqrt{6}+\sqrt{2}}\)

 = \(\sqrt{\left(\sqrt{6}-\sqrt{2}\right).\left(\sqrt{6} +\sqrt{2}\right)}\) = \(\sqrt{6-2}\) = \(\sqrt{4}\) = 2

k mk nha

Dùng tính chất tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+b+c}{b+d+f}\)   ( có b + d + f \(\ne\)0 )

* Trước tiên ta xét trường hợp x + y + z = 0 có :

\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=0\) \(\Rightarrow x=y=z=0\)

* Xét x + y + z = 0 , tính chất tỉ lệ thức :

\(x+y+z=\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{2x+2y+2z}=\frac{1}{2}\)

\(\Rightarrow x+y+z=\frac{1}{2}\)Và \(2x=y+z+1=\frac{1}{2}-x+1\Rightarrow x=\frac{1}{2}\)

                                           \(2y=x+z+1=\frac{1}{2}-y+1\Rightarrow y=\frac{1}{2}\)

                                            \(z=\frac{1}{2}-\left(x+y\right)=\frac{1}{2}-1=\frac{-1}{2}\)

Vậy có cặp \(\left(x,y,z\right)\) thỏa mãn: \(\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2}\right)\)

\(\left(1\right)\Rightarrow\hept{\begin{cases}\sqrt{x}-\sqrt{y}=\frac{1}{\sqrt{z}}-\frac{1}{\sqrt{y}}=\frac{\sqrt{y}-\sqrt{z}}{\sqrt{xy}}\\\sqrt{y}-\sqrt{z}=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{z}}=\frac{\sqrt{z}-\sqrt{x}}{\sqrt{xz}}\\\sqrt{z}-\sqrt{x}=\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\end{cases}\left(2\right)}\)

\(\left(2\right)\Rightarrow\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{y}-\sqrt{z}\right).\left(\sqrt{z}-\sqrt{x}\right)=\frac{\left(\sqrt{y}-\sqrt{z}\right).\left(\sqrt{z}-\sqrt{x}\right).\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{zyzxxy}}\left(3\right)\)\(Từ\left(3\right)\)Ta sẽ chứng minh được rằng \(\orbr{\begin{cases}x=y=z\\x.y.z=1\end{cases}}\)