Question

已知1/x-1/y=3，则分式为2x+3xy-2y/x-2xy-y的值为？？？？？？

Answer 1

解法一:
1)因(1/x) - (1/y ) = 3,可得:(y-x)/xy=3,得:x+3xy-y=0.
2)2x+3xy-2y)/(x-2xy-y)可以进一步分解成:
{x+(x+3xy-y)-Y}/{(x+3xy-y)-5xy}
3)将x+3xy-y=0.代入{x+(x+3xy-y)-Y}/{(x+3xy-y)-5xy}中,
得:(x+0-y)/(0-5xy)=(x-y)/(-5xy),得:-(y-x)/(-5*xy)
同理将1)中的:(y-x)/xy=3 代入:-(y-x)/(-5*xy),
得:-3/(-5)=3/5.

因此(2x+3xy-2y)/(x-2xy-y)=3/5.
解法二:

1)将(2x+3xy-2y)/(x-2xy-y)上下分别都除以xy,得:(2/y+3-2/x)/(1/y-2-1/x)
2)得:(2/y-2/x+3)/(1/y-1/x-2)
因(1/x) - (1/y ) = 3,因此(1/y) - (1/x) = -3
因此得(-3*2+3)/(-3-2)=3/5.

(解法一较正统些水到渠成,方法二,虽简单,但如果思路不对,就可能做不下去的可能.)

Answer 2

上下同时除以xy
易得原式=9