Do \(x+y+z=6\Rightarrow\left(x+y+z\right)^2=6^2\)
\(\Leftrightarrow\left(x+y\right)^2+2.\left(x+y\right)z+z^2=36\)
\(\Leftrightarrow x^2+2xy+y^2+2xz+2yz+z^2=36\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)+2.\left(xy+yz+xz\right)=36\)
\(\Leftrightarrow12+2.\left(xy+yz+zx\right)=36\)
\(\Rightarrow xy+yz+zx=\left(36-12\right):2=12\)
Ta có: \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Mà \(\left(x-y\right)^2;\left(y-z\right)^2;\left(z-x\right)^2\ge0\)
\(\Rightarrow x-y=y-z=z-x=0\)
\(\Rightarrow x=y=z\)
Khi đó: \(x+y+z=6\Leftrightarrow x+x+x=6\)
\(\Leftrightarrow3x=6\Rightarrow x=2\)
\(\Rightarrow x=y=z=2\)
Vậy ______________
