f(x)=a-[1/(2^x+1)],f(x)为奇函数
f(-x)=-f(x)
a-[1/(2^-x+1)]=-a+[1/(2^x+1)]
2a=[1/(2^-x+1)]+[1/(2^x+1)]
=(2^x+1-1)/(2^x+1)+1/(2^x+1)
=1
a=1/2
f(x)=a-[1/(2^x+1)],f(x)为奇函数
f(-x)=-f(x)
a-[1/(2^-x+1)]=-a+[1/(2^x+1)]
2a=[1/(2^-x+1)]+[1/(2^x+1)]
=(2^x+1-1)/(2^x+1)+1/(2^x+1)
=1
a=1/2