若(2x+y-1)的平方+|x-2y-3|(绝对值)=0
所以,2x+y-1 = 0,x-2y-3 = 0
2x+y = 1,x-2y = 3
x = 1, y = -1
(2x+y)(2x-y) -(xy+2)(x-2y- 1)
=(2X1 - 1)(2X1 +1)- (-1X1 + 2)(1 + 2 -1)
= -3
若(2x+y-1)的平方+|x-2y-3|(绝对值)=0
所以,2x+y-1 = 0,x-2y-3 = 0
2x+y = 1,x-2y = 3
x = 1, y = -1
(2x+y)(2x-y) -(xy+2)(x-2y- 1)
=(2X1 - 1)(2X1 +1)- (-1X1 + 2)(1 + 2 -1)
= -3