`-x^2-y^2+x+y+2`
`=-x^2+x-1/4-y^2+y-1/4+2+1/4+1/4`
`=(-x^2+x-1/4)-(y^2-y+1/4)+5/2`
`=-[x^2-2.x. 1/2+(1/2)^2]-[y^2-2.y. 1/2+(1/2)^2]+5/2`
`=-(x-1/2)^2-(y-1/2)^2+5/2`
`=-[(x-1/2)^2+(y-1/2)^2]+5/2`
Có: `(x-1/2)^2\ge0, (y-1/2)^2\ge0 ⇒ (x-1/2)^2+(y-1/2)^2(y-1/2)^2\ge0 `
`⇒-[(x-1/2)^2+(y-1/2)^2]\le0⇒-[(x-1/2)^2+(y-1/2)^2]+5/2\le5/2.`
Dấu bằng xảy ra khi:
$\begin{cases}(x-\dfrac{1}{2})^2=0\\(y-\dfrac{1}{2})^2=0\end{cases}$$⇔\begin{cases}x-\dfrac{1}{2}=0\\y-\dfrac{1}{2}=0\end{cases}$`⇔x=y=1/2.`
Vậy $Min$`=5/2⇔x=y=1/2.`
