令x1=x2+m （m＞0）,即：x1＞x2
lg(1-x1/1+x1) - lg(1-x2/1+x2)
=lg(1-x1/1+x1)/(1-x2/1+x2)
=lg[(1-x1)(1+x2) /（1+x1)(1-x2)]
=lg[(1-x2-m)(1+x2) /（1+x2+m)(1-x2)]
=lg[(1-x2-m)/(1-x2)] - lg[(1+x2+m) /(1+x2)]
=lg[1 - m/(1-x2)] - lg[1+ m /(1+x2)]
当lg[1 - m/(1-x2)] - lg[1+ m /(1+x2)]＞0时增函数,此时1 - m/(1-x2) ＞ 1+ m /(1+x2)
1/(x2 -1) ＞ 1/(x2 +1) 得：x2＞1 故：x＞1时增函数；同理：x＜1时减函数
又1-x/1+x ＞0 得：-1＜x＜1
综上：定义域 -1＜x＜1,y=lg(1-x/1+x)的单调性为减函数.
祝你学习进步!