Sửa đề :
\(8-2x^2-y^2+2xy-4x\)
\(=-\left(x^2-2xy+y^2\right)-\left(x^2+4x+4\right)+12\)
\(=-\left(x-y\right)^2-\left(x+2\right)^2+12\)
Ta có :
\(-\left(x-y\right)^2\le0\forall xy\)
\(-\left(x+2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-y\right)^2-\left(x+2\right)^2+12\le12\forall xy\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x+2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=-2\\x=-2\end{matrix}\right.\)
Vậy \(Max=12\Leftrightarrow x=y=-2\)
8-2x2-y2+2xy-4y= -y2+2y(x-2)-(x-2)2-x2-4x+12
=-(y2-2y(x-2)+(x-2)2)-(x2+4x+4)+16
=-(y-x+2)2-(x+2)2+16 \(\le\) 16, với mọi x,y.
Dấu "=" xảy ra khi : \(\left\{{}\begin{matrix}y-x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-4\\x=-2\end{matrix}\right.\)
Vậy GTLN của biểu thức trên là 16 khi x=-2; y=-4.
\(A=8-2x^2-y^2+2xy-4y\\ =-\left(-8+2x^2+y^2-2xy+4y\right)\\ =-\left(y^2-2y\left(x-2\right)+\left(x-2\right)^2+\left(x^2+4x+4\right)\right)=-\left(\left(y-x+2\right)^2+\left(x+2\right)^2\right)\\ =-\left(y-x+2\right)^2-\left(x+2\right)^2\)
Max A = 0 khi x=-2 ;y=0
3 Người 3 đáp án :V