一道数学计算 1/x-1/y= -3,求(2x+3xy-2y)/(x-2xy-y)

1/x-1/y= -3
---> (x-y)/xy=3
x-y=3xy
(2x+3xy-2y)/(x-2xy-y)
=(6xy+3xy)/(3xy-2xy)
=9xy/xy
=9

(2x+3xy-2y)/(x-2xy-y)
分子分母同除以xy
分子/xy=2/y+3-2/x=3-2(1/x-1/y)=9
分母/xy=1/y-2-1/x=-2-(1/x-1/y)=1
所以原式=9

1/x-1/y= -3
(y-x)/xy=-3
y-x=-3xy ---> x-y=3xy
(2x+3xy-2y)/(x-2xy-y)
=[2(x-y)+3xy] / (x-y-2xy)
=(6xy+3xy) / (3xy-2xy)
=9xy/xy
=9

1/x-1/y= -3通分得(y-x)/xy=-3
y-x=-3xy x-y=3xy
(2x+3xy-2y)/(x-2xy-y)得[2(x-y)+3xy]/[(x-y)-2xy]
把x-y=3xy代入原式得
[2*3xy+3xy]/[3xy-2xy]得9XY/XY=9
2楼的做错了