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.

原式=(x2+y2-x2+2xy-y2+2xy-2y2)÷4y=(4xy-2y2)÷4y=x-

1
2
y=
1
2
(2x-y),
∵2x=y,∴2x-y=0,
则原式=0.