Ta có: \(8-2x^2-y^2+2xy-4y=\left(-y^2+2xy-4y\right)-2x^2+8=\left[-y^2+2y\left(x-2\right)\right]-2x^2+8=-\left[y^2-2y\left(x-2\right)+\left(x-2\right)^2\right]+\left(x-2\right)^2-2x^2+8=-\left[y-\left(x-2\right)\right]^2+x^2-4x+4-2x^2+8=-\left(y-x+2\right)^2-x^2-4x+12=-\left(y-x+2\right)^2-\left(x+2\right)^2+4+12=-\left(y-x+2\right)^2-\left(x+2\right)^2+16\le16\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y-x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-4\\x=-2\end{matrix}\right.\)
Vậy \(GTLN\) của biểu thức đã cho là 16\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-4\end{matrix}\right.\)
Đặt A=\(8-2x^2-y^2+2xy-4y\)
<=>2A=\(16-4x^2-2y^2+4xy-8y\)
<=>2A=\(-(2x-y)^2-(y+4)^2+32\)
Ta có:\(-(2x-y)^2\le0\)
\(-(y+4)^2\le0\)
=> 2A\(\le32\)
=>\(A\le16\)
Dấu "=" xảy ra <=>y=-2,x=-1
