已知x-2y=0,求x−yx+2y÷x2−y2x2+4xy+4y2−2的值.

问题描述:

已知x-2y=0,求

x−y
x+2y
÷
x2y2
x2+4xy+4y2
−2的值.

原式=

x−y
x+2y
(x+2y)2
(x+y)(x−y)
-2
=
x+2y
x+y
-2,
∵x-2y=0,
∴x=2y,
∴原式=
2y+2y
2y+y
-2=-
2
3