已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为

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已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为1/x-

已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为
已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为

已知1/x-1/y=-3,则5x+xy-5y/x-xy-y的值为
1/x-1/y=(y-x)/xy=-3
y=x=-3xy
x-y=3xy
所以原式=[5(x-y)+xy]/(x-y)-xy]
=[5(3xy)+xy]/[(3xy)-xy]
=16xy/2xy
=8

1/x-1/y=-3，
（5x+xy-5y）/(x-xy-y)
=(5/y-5/x+1)/(1/y-1/x-1)
={5(1/y-1/x)+1}/(1/y-1/x-1)
=16/.2
=8

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