∵|x+2|+(y-1/2)²=0
∴x+2=0、y-1/2=0
x=-2、y=1/2
2x²y-2[xy²-3(xy²-2x²+1)]
=2x²y-2[xy²-3xy²+6x²-3]
=2x²y-2xy²+6xy²-12x²+6
=2x²y+4xy²-12x²+6
=2×(-2)²×1/2+4×(-2)×(1/2)²-12×(-2)²+6
=4-2-48+6
=-40
x+2=0,y-1/2=0
x=-2,y=1/2
所以
原式=2x^2y-2xy^2+6(xy^2-2x^2+1)
=2x^2y-2xy^2+6xy^2-12x^2+6
=2x^2y+4xy^2-12x^2+6
=2×4×1/2+4×（-2）×1/4 -12×4+6
=4-2-48+6
=-40
远大教育帮助你，你梦想的起点
x = -2,y = 1/2
由题意知x=-2,y=1/2,将原式化简后代入求值就可以了