已知x分之1-y分之1=3,求分式x-2xy-y分之-搜答案-搜搜考试网

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-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

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-

用x分之3加y分之4乘以（x+y),然后化简,再用均值不等式就出来了……(3/x+4/y)乘（x+y）等于7+3y/x+4x/y然后由于x,y大于0,所以用均值不等式即可……

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

3+9+3=15==

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

1/X-1/Y=5Y-X=5XY,(X-3XY-Y)/(3X+5XY-3Y)=［-(Y-X)-3XY］/［-3(Y-X)+5XY］=(-8XY)/(-10XY)=4/5.

ay/(y-b)

1/x-1/y=(y-x)/xy=2y-x=2xy(3x-5xy-3y)/(-x+3xy+y)={3(x-y)-5xy}/(y-x+3xy)={(-3)*2xy-5xy}/5xy=-11/5

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

解x/y=3∴x=3y∴(x²+xy)/y²=(9y²+3y²)/y²=12y²/y²=12最简公分母(x-1)²(x+

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

x=2分之根号3-根号2y=2分之根号3+根号2xy=1/4y-x=根号2x分之1-y分之1=(y-x)/xy=根号2/(1/4)=4根号2

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

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