若x-y=2016,y-z=2017,则z-x=?
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(y-z)^2+(z-x)^2+(x-y)^2=(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2[(y-z)^2-(y+z-2x)^2]+[(z-x)^2-(x+z-2y)^2]+[(
[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊
有这样的公式:a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3(x+y
设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则(1)x+y-z=kz(2)x-y+z=ky(3)-x+y+z=kx(1)+(2)+(3)得x+y+z=k(x+y+z)∴k=1时,
设x/3=y/4=z/5=m则x=3m,y=4m,z=5m则x+y+z/3x-2y+z=(3m+4m+5m)/(9m-8m+5m)=12/6=2
解:x+1/y=y+1/z=z+1/x.若x=y=z,则(xyz)^2的值并不固定;若x≠y≠z,则:x+1/y=y+1/z,x-y=1/z-1/y=(y-z)/zy,则(x-y)zy=(y-z);(
x:y:z=2:3:4令x=2k,y=3k,z=4k(x+y+z)/(x-2y+3z)=(2k+3k+4k)/(2k-2*3k+3*4k)=9k/(8k)=9/8
2y+z/x=2x+y/z=2z+x/y=k∴2y+z=kx2x+y=kz2z+x=ky三式相加:3(x+y+z)=k(x+y+z)∵x+y+z≠0∴k=3
2y+z/x=2x+y/z=2z+x/y=k法一:发现x,y,z的位置互换依旧为定值,所以x=y=z所以k=(2x+x)/x=3法二:2y+z=kx2x+y=kz2z+x=kx三式相加得:3(x+y+
x=y=z所以k=3
设a=x-y,b=y-z,-a-b=z-x(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方b^2+a^2+(-a-b)^2=(-a-b-
把z=2x+y代入4x=7y+5Z得4x=7y+10x+5yx=-2y,z=-3yx:y:z=-2:1:-3
再答:2016次方,和绝对值里面不可能为负,所以只能所有的都等于0
(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&
=x²(y-z)+y²(z-x)+z²(x-z+z-y)=(y-z)(x²-z²)+(z-x)(y²-z²)=(y-z)(x-z)
设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是:(x+3y-z)/(x-3y+z)?若是,如下:(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3
令a=x/2=y/3=z/4则x=2a,y=3a,z=4a所以原式=(2a+3a+4a)/(2a+3a-4a)=9a/a=9
根据公式(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac公式展开:得到(x^2+y^2+z^2=2xy-2yz-2xz)-(x^2+y^2+z^2-2xy-2yz+2xz)合并同类项
x+3y+10z=0就是x+3y=-10z即2x+6y=-20zA式2x-y-2z=0就是2x-y=2zB式A式-B式得到:(2x+6y)-(2x-y)=-20z-2z即7y=-22z解出y=-22z