Ta có
$-x^2+2xy-2y^2-2x+6y+5\\
=-x^2+2x\left(y-1\right)-2y^2+6y+5$
$=-x^2+2x\left(y-1\right)-y^2+2y-1-y^2+4y-4+10$
$=-\left[x^2-2x\left(y-1\right)+\left(y^2-2y+1\right)\right]-\left(y^2-4y+4\right)+10$
$=-\left[x^2-2x\left(y-1\right)+\left(y-1\right)^2\right]-\left(y-2\right)^2+10\\
=-\left(x-y+1\right)^2-\left(y-2\right)^2+10$
Ta thấy
$-\left(x-y+1\right)^2\le 0\\
-\left(y-2\right)^2\le 0\\
10\ge 10>0$
$\Rightarrow -\left(x-y+1\right)^2-\left(y-2\right)^2+10\le 10$
Dấu bằng xảy ra $\Leftrightarrow x=1;y=2$
Vậy GTLN=10 tại x=1;y=2
