y=x-z与y=z-x的对称
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(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z
[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊
设(y+z)/x=(x+z)/y=(x+y)/z=k;y+z=kx;x+z=ky;y+z=kx;2(x+y+z)=k(x+y+z);k=2或x+y+z=0;所以,(y+z)(x+z)(x+y)/xyz
x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y
x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1
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(y+z)/x=(z+x)/y=(y+x)/z=k则y+z=kx,z+x=ky,y+x=kz三式相加2(x+y+z)=k(x+y+z)故当x+y+z=0时,k=-1,但xy-z不等于0,可知x+y+
令(y+z)/x=(z+x)/y=(x+y)/z=t∴y+z=xt,z+x=yt,x+y=zt三式相加得:2(x+y+z)=(x+y+z)t∴(2-t)(x+y+z)=0∴2-t=0或x+y+z=0若
∵y+z÷x=Z+X÷y=X+Y÷z容易发现x,y,z位置互换也成立∴式子与x,y,z值无关∴x=y=z∴(X+Y-Z)÷(X+Y+z)=x/3x=1/3明教为您解答,请点击[满意答案];如若您有不满
x',y',z'是啥意思?没说是整数还是自然数,或者别的条件?(x-x')+(y+y')+z*z'=16这个式子也没有问题?条件不明确,本题有很多解.后面的两个限制条件没有用.x+y+z=14的自然数
设:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y=k{x+y-z=kz(1){y+z-x=kx(2){z+x-y=ky(3)(1)+(2)+(3)得:(x+y+z)=k(x+y+z)(x
令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0
(x+y)/z=(x+z)/y=(z+y)/xx,y,z等价x=y=z(x+y)(x+z)(z+x)/xyz=8
设x+y-z/z=x-y+z/y=y+z-x/x=k有x+y-z=kzx-y+z=kyy+z-x=kx三式相加得x+y+z=k(x+y+z)k=1得x+y=(k+1)zx+z=(k+1)yy+z=(k
x+y大于等于2倍根号下xy同理x+z大于等于2倍根号下xzz+y大于等于2倍根号下zy所以(x+y)(y+z)(z+x)大于等于8xyz当取到8xyz时分数值最大为1/8此时x=1/3y=1/3z=
就是个三元一次方程组,解出来x=10y=20z=15x+y+z=45
X+Y+Z
求直线L:x+2y+z=1;x+y+2z=4上一点:令z=0,由x+2y=1,x+y=4,得:x=7,y=-3直线L上的点(7,-3,0).这不是唯一的,也可取(0,-2/3,7/3),.直线L的方向
f=x+1f+u=2x+3f+u+c=3x+8f+u+c+k=4x+15f(f,u,c,k)=(x+1)(2x+3)(3x+8)(4x+15)