Giải thích các bước giải:

a.$A=-x^2-4x-y^2+2y=-(x^2+4x+4)-(y^2-2y+1)+5=-(x+2)^2-(y-1)^2+5\le 5$

$\to GTLN_A=5\to x=-2, y=1$

c.$C=-x^2+2x-y^2-4y+6=-(x^2-2x+1)-(y^2+4y+4)+11=-(x-1)^2-(y+2)^2+11\le 11$

$\to GTLN_C=11\to x=1, y=-2$

d.$D=-x^2-2x-y^2+4y+6=-(x+1)^2-(y-2)^2+11 \le 11$

$\to GTLN_D=11\to x=-1, y=2$

e.$E=-x^2-y^2-2(x+y)+3=-(x^2+2x+1)-(y^2+2y+1)+5=-(x+1)^2-(y+1)^2+5\le 5$

$\to GTLN_E=5\to x=-1, y=-1$

f.$-F=x^2-2xy+2y^2+2y+1=(x^2-2xy+y^2)+(y^2+2y+1)=(x-y)^2+(y+1)^2\ge 0$

$\to F\le 0$

$\to GTLN_F=0\to x=y=-1$