∵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...