B = $-x^{2}$ – 2$y^{2}$ – 2xy -2x +2y + 2013 Tìm gtln

Đáp án:

`B=-x^2-2y^2-2xy-2x+2y+2013`

`=-(x^2+2y^2+2xy+2x-2y-2013) `

`=-(x^2+2xy+y^2+2x+2y+y^2-4y-2013)`

`=-[(x+y)^2+2(x+y)+1+y^2-4y+4-2018]`

`=-[(x+y+1)^2+(y-2)^2-2018]`

Ta có: $\left\{\begin{matrix}(x+y+1)^2≥0& \\(y-2)^2≥0& \end{matrix}\right.$

 `-> (x+y+1)^2+(y-2)^2>=0`

`-> (x+y+1)^2+(y-2)^2-2018>=-2018`

`-> -[(x+y+1)^2+(y-2)^2-2018]<=2018`

Dấu “=” xảy ra `<=>` $\left\{\begin{matrix}(x+y+1)^2=0& \\(y-2)^2=0& \end{matrix}\right.$

`->` $\left\{\begin{matrix}x+y+1=0& \\y-2=0& \end{matrix}\right.$

`->` $\left\{\begin{matrix}x+y=-1& \\y=2& \end{matrix}\right.$

`->` $\left\{\begin{matrix}x=-3& \\y=2& \end{matrix}\right.$

Vậy `B_(max)=2018 <=>`$\left\{\begin{matrix}x=-3& \\y=2& \end{matrix}\right.$