f(x)=log2 (1+x)/(1-x)f(-x)=log2 (1-x)/(1+x)=-log2 (1+x)/(1-x)=-f(x)f(x)是奇函数设-1f(x1)-f(x2)=log2 (1+x1)/(1-x1)-log2 (1+x2)/(1-x2)=log2 (1+x1)(1-x2)/(1-x1)(1+x2)=log2 (1+x1-x2-x1x2)/(1-x1+x2-x1x2)因为x1-x2<0,所以0<分子<分母,(1+x1-x2-x1x2)/(1-x1+x2-x1x2)<1所以f(x1)-f(x2)<0f(x)在(-1,1)上单调递增