已知x y分之1

1/x-1/y=3得y-x=3xy∴xy=(y-x)/3则x-xy-y=x-y-(y-x)/3=4(y-x)/35x+xy-5y=5(x-y)+(y-x)/3=14(x-y)/3∴(5x+xy-5y)

1/x-1/y=2y-x=2xyx-y=-2xy(3x+7xy-3y)/(2x-3xy-2y)=[3(x-y)+7xy]/[2(x-y)-3xy]=[-6xy+7xy]/[-4xy-3xy]=xy/[

分子分母同除以xy,分子变为2（1/x+1/y)+3=2x3+3=9分母变为（1/x+1/y)+2=3+2=5答案是9/5

1/X+1/Y=3即x+y=3xy(2x+3xy+2y)/(x-2xy+y)=(2*3xy+3xy)/(3xy-2xy)=9xy/xy=9答案为9

解1/x+1/y=5即(y+x)/xy=5∴y+x=5xy∴(2x+3xy+2y)/(x-2xy+y)=[2(x+y)+3xy]/[(x+y)-2xy]=[2×5xy+3xy]/(5xy-2xy)=(

由1/x-1/y=3得(y-x)/xy=3题中的式子分子分母同时除以xy得(2/y+3-2/x)/(1/y-2-1/x),通分得【2(x-y)/xy+3】/【(x-y)/xy-2】代入得3/5

由　1/X-1/Y=5,两边同乘　XY,得出Y-X=5XY,那么　　分子　-X+XY+Y=6XY分母　2X+7XY-2Y=7XY-2(Y-X)=-3XY　　得此分式的值为－2

1/x-1/y=2(y-x)/xy=2y-x=2xy所以x-y=-2xy原式=[2(x-y)+7xy]分之[(y-x)+xy]=[2(-2xy)+7xy]分之[(2xy)+xy]=3xy]分之3xy=

再问：2/5再答：？

已知x分之1-y分之1=3,则有y-x=3xy(5x+xy-5y)/(x-xy-y)=7/2

1/x-1/y=3x-y=-3xy(2x+3xy-2y)/(x-2xy-y)=[2(x-y)+3xy]/[(x-y)-2xy]=(-3xy)/(-5xy)=3/5

1/x-1/y=3(2x+3xy-2y)/(x-2xy-y)=(2/y+3-2/x)/(1/y-2-1/x)=[3-2(1/x-1/y)]/[-2-(1/x-1/y)]=(3-2*3)/(-2-3)=

(x-3xy-y)分之(2x+3xy-2y)分子分母同时除以xy得=(1/y-3-1/x)分之(2/y+3-2/x)=[-(1/x-1/y)-1]分之(-2(1/x-1/y)+3)=(-3-1)分之(

由x分之1-y分之1=2得到,1/x-1/y=2,即（y-x）/(xy)=2,即x-y=-2xy(1).2x-3xy-2y分之3x+7xy-3y=(2x-3xy-2y)/(3x+7xy-3y)(2)将

直接设X=1,Y=0.5.带入后式得出10

/>1/x-1/y=3(y-x)/(xy)=3y-x=3xyx-y=-3xy(2x-14xy-2y)/(x-2xy-y)=[2(x-y)-14xy]/[(x-y)-2xy]=[2×(-3xy)-14x

(5x+3xy-5y)/(x-2xy-y)=(5/y+3-5/x)/(1/y-2-1/x)分子分母同时除以xy=[-5(1/x-1/y)+3]/[-(1/x-1/y)-2]=(-15+3)/(-3-2

1/x+1/y=-2则(x-xy+y)/(x+xy+y)分子分母同除以xy=(1/x+1/y-1)/(1/x+1/y+1)=(-2-1)/(-2+1)=3

x-y分之xy=-3分之1xy分之(x-y)=-3x-y=-3xyx-2xy-y分之2x+3xy-2y=(-6xy+3xy)/(-3xy-2xy)=3/5

由已知条件可得：x-y=-4xy,（也就是在等式两边乘xy后整理,即可）,代入所求的式子里分母为-4xy+7xy,即3xy；分子为-4xy-2xy,即-6xy,化简得-2