若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值.

若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值.
若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值.

若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值.
x+y=1
x-y=2
(x+2y)(x-2y)-(2x-y)(-y-2x)
=(x+2y)(x-2y)+(2x-y)(y+2x)
=x²-4y²+4x²-y²
=5x²-5y²
=5(x+y)(x-y)
=10

由题意的
x+y=1
x-y=2
x=1.5
y=-0.5
化简所求的答案为10