已知x分之1-y分之1＝4-搜答案-大师作文网

1/x-1/y=2001则(x-xy-y)/(x-y)=2002/2001

∵1/x+1/y=1/6,1/y+1/z=1/9,1/z+1/x=1/15∴（1/x+1/y）+（1/y+1/z）+（1/z+1/x）=1/6+1/9+1/152（1/x+1/y+1/z）=15/90

x/2=y/3=z/4=tx=2ty=3tz=4tx+y-z=1/122t+3t-4t=t=1/12x=1/6y=1/4z=1/3xyz=1/6*1/4/*1/3=1/72

1/x-1/y=4(y-x)/xy=4x-y=-4xy(3X+5xy-3y)/(x-3xy-y)=[3(x-y)+5xy]/[(x-y)-3xy]=(-12xy+5xy)/(-4xy-3xy)=1

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

令2分之x=3分之y=4分之z=k,则：x=2ky=3kz=4kx+y+z=1/12,得：2k+3k+4k=1/129k=1/12k=1/108所以：x=2/108=1/54y=3/108=1/36z

∵（2x+y)/2=(5x+2y)/4=1∴{2x+y=2(1){5x+2y=4(2)(1)==>y=2-2x代入（2）5x+2(2-2x)=4x+4=4,∴x=0,y=2∴(x+2y+1)/(2x-

x分之1-y分之1=4(xy)分之(y-x)=4y-x=4xyx-y=-4xy(x-2xy-y)分之(2x+xy-2y)=[(x-y)-2xy]分之[2(x-y)+xy]=(-4xy-2xy)分之(-

1.令x/2=y/3=z/4=k,则有:x=2ky=3kz=4k所以:(2x+y-z)/(x+y+z)=(2×2k+3k-4k)/(2k+3k+4k)=1/32因为x^2+y^2+2x+4y+4=0即

∵x/4=y/5　∴x:y=4:5　∵y/5=z/3　∴y:z=5:3　∴x:y:z=4:5:3

3/4x=y/5,则x:y=4:15,y/5=z/6,则y:z=5:6,所以:x:y:z=4:15:18

把要求的式子分子分母同时除以xy得(x+2xy+y)/(3x-4xy+3y)=(1/y+2+1/x)/(3/y-4+3/x)=(2+3)(3×3-4)=1

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

(3x-4y)÷(2x+y)=1/2所以2x+y=2(3x-4y)=6x-8y4x=9y所以x/y=9/4

因为：3分之x+1=4分之y+3=5分之x+y,所以：（x+1）/3=（y+3）/4=(x+y)/54x+4=3y+94x=3y+55x+5=3x+3y2x=3y-5x=5y=5则x+2y+3分之3x

因为x＋y/2=3x＋4y/5=1,所以x＋y=2,3x＋4y=5所以解方程组得x=3,带入x＋5y＋6/3x－y＋1,得4/11

x/3=y/4=z/5x：Y：Z=3；4;5

原式等价于10(x+1)=5(y+3)=4(x+y)10(x+1)=5(y+3)可得2x-y=15(y+3)=4(x+y)可得4x-y=15解得x=7y=13x+2y+3=363x+2y+1=48两式

1/x+1/y=-1/(x+y)两边同时乘x+y得1+y/x+1+x/y=-1所以y/x+x/y=-3