1/x+1/y＋1/z=1.x,y,z,属于不同自然数 y,

已知 x/(y+z)+y/(z+x)+z/(x+y)=1求 (x*x)/(y+z)+(y*y)/(x+z)+(z*z)/(x+y)=? 1/x+1/y＋1/z=1.x,y,z,属于不同自然数 y, 1.设有比例式：x/(y+z)=y/(x+z)=z/(x+y),有比例性质,得x/(y+z)=y/(x+z)=z/(x+y)=（x+y+z)/[(y+z)+(x+z)+(x+y)]=0.5x/(y+z)=y/(x+z)=(x-y）/[（y+z）-（x+z)]=-1,由此可得0.5=-1试分析推理产生错误的原因 {x+y-z=11 y+z-x=5 z+x-y=1 {x+y-z=11,x-y-z=-5,x-y+z=1 1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z) (1)X+Y=15,Y+Z=-6,Z+X=7.(2)X+Y=-14,Y+Z=-7,Z+X=9.(3)X+Y=0,Y+Z=-1,Z+X=-1. X/Y+Z=Y/X+Z+1=Z/X+Y-1=X+Y+Z,求x+y+z等于多少 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值 x+y+z=6 x+2y+3z=14 y+1=z 1.{x:y:z=3:4:5 2x+3y-5z=-142.{x-y-z=-1 3x+5y+7z=11 4x-y+2z=-13.{x+y=1 x+z=0 2x+y+2z=44.{x-y+z=-3 2x-y-z=-2 -3x+y+z=-15.{x-y+z=-3 3x+y-4z=6 2x+y-z=36.{x-2y+5z=16 2x+y-5z=7 2x-y+z=38 有谁能帮俺解释一下++x||++y&&++z是什么意思啊?main(){int x,y,z (x,y,z);x=y=z=1; ++x||++y&&++z; printf(x=%d,y=%d,z=%d ,x,y,z) } 已知实数X.Y.Z满足(Y+Z)分之X+(Z+X)分之Y+(X+Y)分之Z=1,则(Y+Z)分之X平方+(Z+X)分之Y平方+(X+Y)分之Z平方的值为（ ） 挺急的~已知Z分之X+Y-Z=y分之X-Y+Z=X分之-X+Y+Z且XYZ分之(X+Y)(X+Z)(Y+Z)=-1.x+y+z的值是还想问下这错在哪儿啊~设X+Y-Z/Z=X-Y+Z/Y=-X+Y+Z/X=K则X+Y=Z(K+1),X+Z=Y(K+1),Y+Z=X(K+1)因为XYZ分之(X+Y)(X+Z)(Y+Z)=-1所以(K+1)三次方=- 若|x+3|+|y-2|+|2×z+1|=0求(x×z-y×z)(y-x+z)的值 x,y,z都>0.且x+y+z=1.求证1/x+4/x+9/z>=36 f(x,y,z)=In{[x^α+z^(1-α-β)]/y^β}+x/z+(z/x)^β,求 ∂f(x,y,z)/∂y