题目应为：f(x) = log2(1+x) - log2(1-x)
f(x) = log2(1+x) - log2(1-x) = log2 [(1+x)/(1-x) ]
f(-x) = log2 [(1-x)/(1+x) ] = -l og2 [(1+x)/(1-x) ] = -f(x),奇函数
f(a) = log2 [(1+a)/(1-a) ]
f(b) = log2 [(1+b)/(1-b) ]
f[(a+b)/(1+ab)]=2
log2【{1+(a+b)/(1+ab)} / {1-(a+b)/(1+ab)} 】= 2
log2【(1+ab+a+b) / (1+ab-a-b) 】 = 2
log2【(1+a)(1+b)/[(1-a)(1-b)] 】= 2
log2【(1+a)/(1-a) * (1+b)/(1-b) 】= 2
log2【(1+a)/(1-a)】+ log2【(1+b)/(1-b) 】= 2
f(a) + f(b) = 2 .（1）
f(-b)= 1/2
f(b) = -f(-b) = -1/2 .(2)
f(a) = 2-f(b) = 2-(-1/2) = 5/2