已知函数f（x）=a+(2/x-1) g(x)=f(2^x)

已知函数f（x）=a+(2/x-1) g(x)=f(2^x)
g(x)=a+2/(2^x-1)
奇函数,则f(-x)=-f(x)
x=1 f(1)=a+2/(2^1-1)=a+2
x=-1 f(-1)=a+2/(-1/2)=a-4
f(-1)+f(1)=0
所以 a+2+a-4=0
2a-2=0
a=1