\(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\Rightarrow x=y=z\)
Ta lại có : \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)
\(\Rightarrow3x^{2009}=3^{2010}\Rightarrow x^{2009}=3^{2009}\Rightarrow x=3\)
\(\Rightarrow x=y=z=3\)
Vậy .............