设函数f(x)=(a·2的x次方-1）/1+2的x次方（a∈R),均满足f(-x)=-f(x).

f（x）=（a*2^x-1)/(1+2^x)
f(-x)=(a*2^(-x)-1)/(1+2^(-x))=(a-2^x)/(2^x+1)
f(x)是R上的奇函数,f(x)=-f(-x)
（a*2^x-1)/(1+2^x)=-(a-2^x)/(2^x+1)
a*2^x-1=-a+2^x
(a-1)*(2^x+1)=0
a=1
f(x)=(2^x-1)/(2^x+1)
2)
f(x)=(2^x-1)/(2^x+1)=1-2/(2^x+1)1
-2/(2^x+1)>-2
f(x)=1-2/(2^x+1)>1-2=-1
-1