由已知|x|+(y-2)²=1,且x,y都为整数得
(1)、|x|=1,(y-2)²=0则x=1,y=2,x-y(x+y)+xyy=-1
或x=-1,y=2,x-y(x+y)+xyy=-7
(2)|x|=0,(y-2)²=1则x=0,y=3,x-y(x+y)+xyy=-9
或x=0,y=1,x-y(x+y)+xyy=-1
由已知|x|+(y-2)²=1,且x,y都为整数得
(1)、|x|=1,(y-2)²=0则x=1,y=2,x-y(x+y)+xyy=-1
或x=-1,y=2,x-y(x+y)+xyy=-7
(2)|x|=0,(y-2)²=1则x=0,y=3,x-y(x+y)+xyy=-9
或x=0,y=1,x-y(x+y)+xyy=-1