若实数x,y满足（x+2y-2)(3x+2y+2)+2(x2+4)=0,求x+y的值.

∵(x+2y-2)(3x+2y+2)+2(x^2+4)
=（2x+2y-x-2)(2x+2y+x+2)+2(x^2+4)
= [(2x+2y)-(x+2)][(2x+2y)+(x+2)]+2(x^2+4)
= (2x+2y)^2-(x+2)^2+2x^2+8
= (2x+2y)^2-x^2-4x-4+2x^2+8
= (2x+2y)^2+(x^2-4x+4)
= (2x+2y)^2+(x-2)^2
= 0
又∵(2x+2y)^2≥0,(x-2)^2≥0,
∴ (2x+2y)^2=(x-2)^2=0
∴2(x+y)=2x+2y=0
∴x+y=0