f(x)=lg{（2/x+1）-1}=lg[(2-x-1)/(x+1)]=lg[(1-x)/(x+1)]f(-x)=lg[(1-(-x))/(-x+1)]=lg[(x+1)/(1-x)]=lg[(1-x)/(1+x)]^-1=-lg[(1-x)/(x+1)]-f(-x)=lg[(1-x)/(x+1)]所以f(x)=-f(-x)所以函数y=lg{（2/x+1）-1}是奇函...