\(B=x^2+5y^2-4xy-5y+6=x^2-4xy+4y^2+y^2-2.\dfrac{5}{2}y+\dfrac{25}{4}-\dfrac{1}{4}\)\(=\left(x-2y\right)^2+\left(y-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\)
Do \(\left(x-2y\right)^2\)\(\ge0\left(\forall x;y\right)\)
\(\left(y-\dfrac{5}{2}\right)^2\ge0\left(\forall y\right)\)
\(\Rightarrow\left(x-2y\right)^2+\left(y-\dfrac{5}{2}\right)^2\ge0\left(\forall x;y\right)\)
\(\Rightarrow\)\(\left(x-2y\right)^2+\left(y-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\left(\forall x;y\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}(x-2y)^2=0\\\left(y-\dfrac{5}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(MinB=\dfrac{-1}{4}\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{5}{2}\end{matrix}\right.\)