Question

已知函数f(x)=lg(1-x)-lg(1+x)求 （1）奇偶性 （2）单调性（3）f(a)+f(b)=f((a+b)/(1+ab)

(2)设-1<x1<x2<1
f(x2)-f(x1)
=lg[(1-x2)/(1+x2)]-lg[(1-x1)/(1+x1)]
=lg{[(1-x2)(1+x1)/[(1+x2)(1-x1)]}
=lg[(1+x1-x2-x1x2)/(1+x2-x1-x1x2)]
=lg[(1+x1-x2-x1x2)/(1+x2-x1-x1x2)]—————————————怎么推倒下一步的
=lg[1-2(x2-x1)/(1+x2-x1-x1x2)]
<lg1=0
即：f(x2)<f(x1)
单调递减！

Answer 1

分子上凑一个1+x2-x1-x1x2，
则1+x1-x2-x1x2=1+x2-x1-x1x2-2(x2-x1)

Answer 2

(1+x1-x2-x1x2)/(1+x2-x1-x1x2)
=[1+x2-x1-x1x2-2(x2-x1)]/(1+x2-x1-x1x2)
=1-2(x2-x1)/(1+x2-x1-x1x2)