(x)关于原点对称
f(x)=-f(-x)
loga[(1-mx)/(x-1)]=-loga[(1+mx)/(-x-1)]
(1-mx)/(x-1)=(-x-1)/(1+mx)
(1-mx)(1+mx)=-(x+1)(x-1)
1-(mx)^2=1-x^2
(m^2-1)x^2=0
m=±1
因为：m=1时,[(1-mx)/(x-1)]=-1