Question

已知1/x-1/y=3,求分式(2x+3xy-2y)/(x-2xy-y)的值

这道题该怎么解啊？要详细的！

Answer 1

将分式(2x+3xy-2y)/(x-2xy-y)的分子分母同除以xy(因为xy不为零),
(2x+3xy-2y)/(x-2xy-y)
=（2/y+3-2/x)/(1/y-2-1/x）
=[3-2(1/x-1/y)]/[-2-(1/x-1/y)]
=(3-2*3)/(-2-3)
=3/5

Answer 2

解：
1/x-1/y=3
y-x=3xy

(2x+3xy-2y)/(x-2xy-y)=(2x+y-x-2y)/(x-2/3(y-x)-y)
=(x-y)/(3/5)(x-y)
=5/3

Answer 3

1/x-1/y=3
(y-x)/xy=3
y-x=3xy
x-y=-3xy
所以(2x+3xy-2y)/(x-2xy-y)
=[2(x-y)+3xy]/[(x-y)-2xy]
=-3xy/-5xy
=3/5