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

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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
因为|x+2y-1|>=0 (y+2)^2>=0
所以|x+2y-1|=(y+2)^2=0
x+2y-1=y+2=0
x+2y=1 y=-2
x=-2y+1=-2*(-2)+1=5
(2x-y)²-2(2x-y)(x+2y)+(x+2y)²
=[(2x-y)-(x+2y)]^2
=(2x-y-x-2y)^2
=(x-3y)^2
=[5-3*(-2)]^2
=(5+6)^2
=121

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