解
x²+2y²-2xy+4y+4=0吧
(x²-2xy+y²)+(y²+4y+4)=0
(x-y)²+(y+2)²=0
∴x-y=0,y+2=0
∴x=-2,y=-2
∴[(xy+2)(xy-2)-2x²y²+4]
=(x²y²-4-2x²y²+4)
=-x²y²
=-4×4
=-16