∴y=loga
1+x |
1-x |
即ay=
1+x |
1-x |
∴ay-1=x(ay+1)
即x=
ey-1 |
ey+1 |
ax-1 |
ax+1 |
(2)f-1(1)=
1 |
3 |
∴
1 |
3 |
a-1 |
a+1 |
∴f-1(x)=
2x-1 |
2x+1 |
∴2x(1-m)<1+m;--------------------------(6分)
①當m≥1時,不等式解集為x∈R;--------------------------(8分)
②當-1<m<1時,得2x<
1+m |
1-m |
1+m |
1-m |
③當m≤-1時,x∈φ--------------------------(12分)