高1数学题,求解!

g(x)=ln(1+x)-ln(1-x)
=ln[(1+x)/(1-x)]
f(x)=2x+g(x)
=2x+ln[(1+x)/(1-x)]
则f(-x)
=-2x+ln[(1-x)/(1+x)]
=-2x-ln[(1+x)/(1-x)]
=-f(x)
(-1<X<1)
任取X1,X2属于(-1,1),且X1>X2
f(x1)-f(x2)
=2(x1-x2)+{ln[(1+x1)/(1-x1)]-ln[(1+x2)/(1-x2)]
=2(x1-x2)+ln{[(1-x1x2)+(x1-x2)]/[(1-x1x2)-(x1-x2)]}
由-1<x2<x1<1
则有: 1-x1x2>0
x1-x2>0
(1-x1x2)+(x1-x2)>(1-x1x2)-(x1-x2)
[(1-x1x2)+(x1-x2)]/[(1-x1x2)-(x1-x2)]>1
则ln{[(1-x1x2)+(x1-x2)]/[(1-x1x2)-(x1-x2)]}>0,2(x1-x2)>0
则f(x1)-f(x2)>0
即1>x1>x2>-1时,f(x1)>f(x2)

f(t)+f(t^2-1)<0
f(t^2-1)<-f(t)
f(t^2-1)<f(-t)

则有:1+t>0-----(1)
1-t>0-----(2)
t^2-1<-t-----(3)
解得：-1<t<(根号5-1)/2

这题用到代入法,自己代进去吧!

哈哈!加油吧 ^-^ !

太久了，忘了

g(x)=ln(1+x)-ln(1-x)

很容易得到定义域是(-1,0)并上（0,1)

f(x)=2x+g(