已知 1/x-1/y=2,求2x+5xy-2y/x-2xy-y

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已知 1/x-1/y=2,求2x+5xy-2y/x-2xy-y

已知 1/x-1/y=2,求2x+5xy-2y/x-2xy-y
已知 1/x-1/y=2,求2x+5xy-2y/x-2xy-y

已知 1/x-1/y=2,求2x+5xy-2y/x-2xy-y
1/x-1/y=2,
2x+5xy-2y/x-2xy-y
分子分母同时除以xy得
(2/y+5-2/x)/(1/y-2-1/x)
=[2(1/y-1/x)+5]/[(1/y-1/x)-2]
=[2x(-2)+5]/(-2-2)
=-1/4

1/x-1/y=2
(y-x)/xy=2
x-y=-2xy
2x+5xy-2y/x-2xy-y
=[2(x-y)+5xy]/[(x-y)-2xy]
=(-4xy+5xy)/(-2xy-2xy)
=xy/(-4xy)
=-1/4

由 1/x-1/y=2得：
(y-x)/xy=2
即y-x=2xy
故
（2x+5xy-2y）/（x-2xy-y）
=[2(x-y)+5xy]/[(x-y)-2xy]
=(-4xy+5xy)/(-2xy-2xy)
=xy/-4xy
=-1/4

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