问题描述: 设f(x)=log 1个回答 分类:数学 2014-11-05 问题解答: 我来补答 ∵f(x)=log121−axx−1为奇函数,∴f(-x)=-f(x),即f(x)+f(-x)=0,则log121−axx−1+log121+ax−x−1=log12(1−axx−1•1+ax−x−1)=0,即a2x2−1x2−1=1,即a2x2-1=x2-1,即a2=1,解得a=1或a=-1,当a=1时,f(x)=log121−axx−1=log12(−1)不成立,当a=-1时,f(x)=log121+xx−1满足条件,故答案为:-1 展开全文阅读