x^2+4y^2-2xy+12y+12=x^2-2xy+y^2+3(y^2+4y+4)=(x-y)^2+3(y+2)^2=0所以：x-y=0,y-2=0即x=y=-2则x^y=(-2)^(-2)=1/(-2)^2=1/4
(x-y)²+3(y+2)²=0，x=y，y=-2，x^y=1/4.
