已知:x分之1-y分之1=3,求x-xy-y分之5x xy-5y的值
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1/x-1/y=2001则(x-xy-y)/(x-y)=2002/2001
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
得xy分之y-x=3得y-x=3xy带入得-9xy-5xy分之-3xy-2xy得14分之5..求采纳.谢谢
∵(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-
∵3x-2y分之5x+y=2分之1∴(5x+y)×2=(3x-2y)×1【两边同乘以2(3x-2y),即交叉相乘】∴10x+2y=3x-2y∴10x-3x=2y+2y∴7x=4y.(1)原式x+y分之
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
∵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
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
(x+y)/2-(x-y)/3=1等式两边同时乘以6,得3(x+y)-2(x-y)=63x+3y-2x+2y=6x+5y=65y=6-xy=(6-x)/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