\(x^2+10y^2+2y-6xy+7=\left(x^2-6xy+9y^2\right)+\left(y^2+2y+1\right)+6=\left(x-3y\right)^2+\left(y+1\right)^2+6\ge6\)\(min=6\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)
\(x^2+10y^2+2y-6xy+7\\ =\left(x^2-6xy+9y^2\right)+\left(y^2+2y+1\right)+6\\ =\left(x-3y\right)^2+\left(y+1\right)^2+6\ge6\)
\(BT_{max}=6\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)
\(x^2+10y^2+2y-6xy+7\)
\(=\left(x^2-2.x.3y+\left(3y^2\right)\right)+\left(y^2+2.y.1+1\right)+6\)
\(=\left(x-3y\right)^2+\left(y+1\right)^2+6\)
Vì \(\left(x-3y\right)^2\text{≥}0\); \(\left(y+1\right)^2\text{≥}0\)
⇒\(\left(x-3y\right)^2+\left(y+1\right)^2+6\text{≥}6\)
Min =6 ⇔y+1=0⇒y=-1
⇔x-3y=0⇒x=-3
