Question

已知1/X-1/y=5,求分式(y+xy-x)/(2x+7xy-2y)的值

Answer 1

(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)
=(1/x+1-1/y)/(2/y+7-2/x)
=6/(-10+7)
=-2

Answer 2

∵1/X-1/y=5，∴（1/X-1/y）乘以xy则有y-x=5xy。
将y-x=5xy代入分式，
则有(y+xy-x)/(2x+7xy-2y)=（y-x+xy）/[7xy-2(y-x)]=（5xy+xy）/(7xy-10xy)
将xy消掉，答案是 -2。

Answer 3

解：因为1/x-1/y=5 ，所以y-x=5xy (两边各乘以xy)
将上式代入，得 (5xy+xy)/[(2x-2y)+7xy]
= 6xy/[7xy-2(y-x)]
= 6y/[7xy-2(5xy)]
= 6xy/(-3xy)
= -2

Answer 4

(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)
=(1/x+1-1/y)/(2/y+7-2/x)
=6/(-10+7)
=-2

Answer 5

由1/X-1/y=5得（y-x）/xy=5 ,y-x=5xy
(y+xy-x)/(2x+7xy-2y)=(y-x+xy)/[-(2y-2x)-7xy]=(5xy+xy)/(-10xy-7xy)=6xy/(-17xy)=-6/17