已知x分之1-y分之1=4
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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
3)把①式乘以10,②式乘以6,得:25x+2y=-402x+y=1由下式得y=1-2x,代入上式得25x+2-4x=-4021x=-42x=-2则y=1+4=54)把①式乘以6,②式乘以2得3x-y
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-
由4x+y-4x-6y+10=0得(4x-4x+1)+(y-6y+9)=0(2x-1)+(y-3)=0(2x-1)=0(y-3)=0x=1/2y=3代入所求算式即可求出结果(不好意思,看不清你的所求算
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
ay/(y-b)
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
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
x/4+3y/2=11)y=2/3(1-x/4)=2/3-x/62)x=4(1-3y/2)=4-6y3)x=-2,y=2/3+2/6=1
设1/(x-y)=a,1/(x+y)=b则原方程组变成:2a+b=1/4b-a=1解得:a=-1/4,b=3/4则1/(x-y)=-1/41/(x+y)=3/4解得:x=-4/3,y=8/3