(x+y-z)/z=(y+z-x)/x=(z+x-y)/y 求(x+y)(y+z)(z+x)/xyz

(x+y-z)/z=(y+z-x)/x=(z+x-y)/y 求(x+y)(y+z)(z+x)/xyz

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+y+z)(k-1)=0k=1或x+y+z=0当k=1时{x+y-z=z{y+z-x=x{z+x-y=y解得x=y=z (x+y)(y+z)(z+x)/xyz=8 当x+y+z=0时(x+y)(y+z)(z+x)/xyz=(-z)(-x)(-y)/xyz=-1======以下答案可供参考======供参考答案1:(x+y-z)/z=（x+y）/z-1=(y+z-x)/x=(y+z)/x-1=(z+x-y)/y=(z+x)/y-1(x+y)/z=(y+z)/x=(z+x)/y所以(x+y)(y+z)(z+x)/xyz=1供参考答案2:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y等价于(x+y)/z-1=(y+z)/x-1=(z+x)/y-1即(x+y)/z=(y+z)/x=(z+x)/y所以(x+y)(y+z)(z+x)/xyz =[(x+y)/z][(y+z)/x][=(z+x)/y]=[(z+x)/y]^3供参考答案3:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y =kx+y-z=kzy+z-x=kxz+x-y=ky 三个式子相加得k=1 所以x=y=z (x+y)(y+z)(z+x)/xyz=8供参考答案4:。。首先x+y+z=0，大家都推出来了，然后(x+y)(y+z)(z+x)/xyz分解后=（x+y）/z +(y+z)/x +(x+z)/y +2 =(x+y+z)/z -1 +(x+y+z)/x -1 +(x+y+z)/y -1 +2 = -1供参考答案5:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y(x+y)/z-1=(y+z)/x-1=(z+x)/y-1x/z+y/z=y/x+z/x=z/y+x/yx/z+y/z=y/x+z/xy=(z/x-x/z)/(1/z-1/x)=(z^2-x^2)/(x-z)=-(x+z)同理，x=-(y+z),z=-(x+y)(x+y)(y+z)(z+x)/xyz=-1供参考答案6:解法一：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+y+z)(k-1)=0k=1或x+y+z=0当k=1时{x+y-z=z{y+z-x=x{z+x-y=y解得x=y=z (x+y)(y+z)(z+x)/xyz=8 当x+y+z=0时(x+y)(y+z)(z+x)/xyz=(-z)(-x)(-y)/xyz=-1解法二：(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

我明天再问问老师，叫他解释下这个问题