解：设(y+z)/x=(z+x)/y=(x+y)/z=k∵(y+z)/x=k∴y+z=kx同理，x+z=ky,x+y=kz三式相加，得(y+z)+(x+z)+(x+y)=kx+ky+kz即2(x+y+z)=k(x+y+z)∵x+y+z≠0，两边约去x+y+z，得k=2∵x+y=kz∴x+y=2z∴(x+y-z)/(x+y+z)=(2z-z)/(2z+z)=z/3z=1/3
