已知1/X-1/y=5,求分式(y+xy-x)/(2x+7xy-2y)的值
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已知1/X-1/y=5,求分式(y+xy-x)/(2x+7xy-2y)的值
(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)
=(1/x+1-1/y)/(2/y+7-2/x)
=6/(-10+7)
=-2
(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)
=(1/x+1-1/y)/(2/y+7-2/x)
=6/(-10+7)
=-2
因为1/x-1/y=5 ,所以y-x=5xy (两边各乘以xy)
将上式代入,得 (5xy+xy)/[(2x-2y)+7xy]
= 6xy/[7xy-2(y-x)]
= 6y/[7xy-2(5xy)]
= 6xy/(-3xy)
= -2
∵1/X-1/y=5,∴(1/X-1/y)乘以xy则有y-x=5xy。
将y-x=5xy代入分式,
则有(y+xy-x)/(2x+7xy-2y)=(y-x+xy)/[7xy-2(y-x)]=(5xy+xy)/(7xy-10xy)
将xy消掉,答案是 -2。
由1/X-1/y=5得(y-x)/xy=5 ,y-x=5xy
(y+xy-x)/(2x+7xy-2y)=(y-x+xy)/[-(2y-2x)-7xy]=(5xy+xy)/(-10xy-7xy)=6xy/(-17xy)=-6/17
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