若X/1-y/1=3 求分式:x-2xy-y/2x+3xy-2y
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若X/1-y/1=3 求分式:x-2xy-y/2x+3xy-2y
中间的 斜线就是分数线,- - ,//
中间的 斜线就是分数线,- - ,//
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.)
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