\(x^2-xy+y^2-4=0\)
\(\Leftrightarrow x\left(x-y\right)+\left(y-2\right)\left(y+2\right)=0\)
Để \(x\left(x-y\right)+\left(y-2\right)\left(y+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x-y\right)=0\\\left(y-2\right)\left(y+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy x = y = 2 hoặc x = y = -2