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+z不等于0故x+y+z不等于0,则k=2[(y+z)(z+x)(x+y)]/xyz=(y+z)/x*(z+x)/y*(x+y)/z=k^3=8...