若|x+2y-1|+y²+4y+4=0,求（2x-y）²-2（2x-y)(x+2y)+(x+2y)²的值

若|x+2y-1|+y²+4y+4=0,求（2x-y）²-2（2x-y)(x+2y)+(x+2y)²的值
若|x+2y-1|+y²+4y+4=0,求（2x-y）²-2（2x-y)(x+2y)+(x+2y)²的值

若|x+2y-1|+y²+4y+4=0,求（2x-y）²-2（2x-y)(x+2y)+(x+2y)²的值
∵|x+2y-1|+y²+4y+4=0
∴|x+2y-1|+(y+2)²=0
∴x=5,y=-2
(2x-y)²-2(2x-y)(x+2y)+(x+2y)²
=[(2x-y)-(x+2y)]²
=(x-3y)²
=[5-3×(-2)]²
=11²
=121