\(C=9x^2+y^2-2xy-8x+10\)
\(=\left(x^2-2xy+y^2\right)+\left(8x^2-8x+2\right)+8\)
\(=\left(x-y\right)^2+8\left(x-\dfrac{1}{2}\right)^2+8\)
Do : \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\8\left(x-\dfrac{1}{2}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-y\right)^2+8\left(x-\dfrac{1}{2}\right)^2+8\ge8\)
Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy GTNN của C là 8 khi \(x=y=\dfrac{1}{2}\)