\(C=x^2-xy+y^2-2x-2y\Leftrightarrow2C=2x^2-2xy+2y^2-4x-4y\)
\(\Leftrightarrow2C=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)
\(\Leftrightarrow2C=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\)
\(\Leftrightarrow C=\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8}{2}\)
\(\Leftrightarrow C=\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2}{2}-4\)
Vì \(\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2}{2}\ge0\)
\(\Rightarrow\)\(\frac{\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2}{2}-4\ge-4\)
Hay \(C\ge-4\)
Vậy \(GTNN\) của \(C=-4\Leftrightarrow\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)
