Đáp án:
`P=2x-2xy-2x^2-y^2`
`=-(2x^2+y^2+2xy-2x)`
`=-(x^2+y^2+2xy+x^2-2x)`
`=-[(x+y)^2 +(x^2-2x+1)-1]`
`=-[(x+y)^2 +(x-1)^2-1]`
`=-(x+y)^2-(x-1)^2+1<=1`
Dấu "=" xảy ra `<=>`
$\left\{\begin{matrix}x+y=0& \\x-1=0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x=-y& \\x=1& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}y=-1& \\x=1& \end{matrix}\right.$
Vậy `P_(max)=1 <=> x=1; y=-1`
