2 |
1−x |
∴f(x)=lg(
2 |
1−x |
2 |
1−x |
1+x |
1−x |
2 |
1−x |
设-1<x1<x2<1,t1−t2=
2 |
1−x1 |
2 |
1−x2 |
2(x1−x2) |
(1−x1)(1−x2) |
∴t1<t2,∴lgt1<lgt2∴f(x1)<f(x2),故y=f(x)在(-1,1)上是单调增函数
又∵f(x)是奇函数,∴f(0)=0,则f(x)<0化为
1+x |
1−x |
1+x |
1−x |
2x |
1−x |
故解集为:(-1,0).
2 |
1−x |
2 |
1−x |
2 |
1−x |
2 |
1−x |
1+x |
1−x |
2 |
1−x |
2 |
1−x1 |
2 |
1−x2 |
2(x1−x2) |
(1−x1)(1−x2) |
1+x |
1−x |
1+x |
1−x |
2x |
1−x |