已知2x=y,求代数式[(x2+y2)-(x-y)2+2y(x-y)]÷(4y)的值.
问题描述:
已知2x=y,求代数式[(x2+y2)-(x-y)2+2y(x-y)]÷(4y)的值.
答
原式=(x2+y2-x2+2xy-y2+2xy-2y2)÷4y=(4xy-2y2)÷4y=x-
y=1 2
(2x-y),1 2
∵2x=y,∴2x-y=0,
则原式=0.