x=y=z=0 k=x ||y &&z

(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

∵（y＋z）/x＝（z＋x）/y＝（x＋y）/z＝k,∴由等比定理,有：k＝［（y＋z）＋（z＋x）＋（x＋y）］/（x＋y＋z）＝2.∴满足条件的k值是2.

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

∵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

已知x,y,z满足y+z/x=z+x/y=x+y/z=k,求k的值解:∵y+z/x=z+x/y=x+y/z=k∴y+z/x=kz+x/y=kx+y/z=k∴kx=y+zky=x+zkz=x+y∴kx+

两种情况：当X+Y+Z=0时：K=(Y+Z)/X=(-X)/X=-1当X+Y+Z不=0时：K=[(Y+Z)+(Z+Y)+(X+Y)]/(X+Y+Z)=2

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

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

1)如果X+Y+Z不=0.y+z/x=z+x/y=x+y/z=k前面3式分子,分母相加.2（x+y+z)/(x+y+z)=kk=22)如果X+Y+Z=0.y+z=-x,x+y=-z,z+x=-yk=-

已知x,y,z满足y+z/x=z+x/y=x+y/z=k,求k的值解:∵y+z/x=z+x/y=x+y/z=k∴y+z/x=kz+x/y=kx+y/z=k∴kx=y+zky=x+zkz=x+y∴kx+

(y+z)/x=（z+x)/y=（x+y)/z=k所以：y+z=kx,z+x=ky,x+y=kz上述三个等式相加,得到：2(x+y+z)=k(x+y+z)得到：k=2

x/(y+z)=y/(x+z)=z/(x+y)=kx=k(y+z)y=k(x+z)z=k(x+y)x+y+z=k(2x+2y2+z)=2k(x+y+z)当x+y+z0时k=1/2当x+y+z=0时k为

分两类情况讨论：(1)根据等比性质：y+z/x=x+z/y=x+y/z=2(x+y+z)/x+y+z=2这种情况是在x+y+z≠0的前提下.(2)当x+y+z=0时,即y+z=-x那么y+z/x=(-

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=3ky=4kz/x=5/3相加为7k+5/3

k(x+y+z)=2(x+y+z)算到这一步时,要讨论当x+y+z不等于0时,k=2当x+y+z=0时,y+z=-x,y+z/x=-1另一个值是-1

k(x+y+z)=2(x+y+z)算到这一步时,要讨论当x+y+z不等于0时,k=2当x+y+z=0时,y+z=-x,y+z/x=-1另一个值是-1

由幂平均不等式得[(x^(k+1)+y^(k+1)+z^(k+1))/3]^[1/(k+1)]≥[(x^k+y^k+z^k)/3]^(1/k)=(1/3)^(1/k),故x^(k+1)+y^(k+1)