已知x分之1减去y分之1等于2 求2x-3xy-2y分之3x+7xy-3y等于多少
问题描述：
已知x分之1减去y分之1等于2 求2x-3xy-2y分之3x+7xy-3y等于多少
答
答案: -1/7
答
因为
1/x - 1/y = 2，
即 (y-x)/xy = 2
所以y - x = 2xy
所以
(3x+7xy-3y) / (2x-3xy-2y)
= 1 + (x + 10xy - y) / (2x - 3xy - 2y)
= 1 + 8xy / -7xy
= 1 - 8/7 = -1/7
答
1/x-1/y=2
y-x=2xy
x-y=-2xy
(3x+7xy-3y)/(2x-3xy-2y)
=[3(x-y)+7xy]/[2(x-y)-3xy]
=[-6xy+7xy]/[-4xy-3xy]
=xy/[-7xy]
=-1/7
答
-1/7.
1/x-1/y=2可得x-y=-2xy;带入后式可得其结果为-1/7。