已知函数f(x）=log以2为底（1+x)/(1-x)求证;f(x1）+f(x2)=f[(a+b)/(1+x1x2)已知函数f(x）=log以2为底（1+x)/(1-x)求证;f(x1）+f(x2)=f[(a+b)/(1+x1x2)若f[(a+b)/(1+ab)=1,F(-b)=1/2,求f(b)的值

Question

已知函数f(x）=log以2为底（1+x)/(1-x) 求证;f(x1）+f(x2)=f[(a+b)/(1+x1x2)
已知函数f(x）=log以2为底（1+x)/(1-x) 求证;f(x1）+f(x2)=f[(a+b)/(1+x1x2)
若f[(a+b)/(1+ab)=1 ,F(-b)=1/2,求f(b)的值

Answer 1

f(x1)+f(x2)
=log2(1+x1)/(1-x1)+log2(1+x2)/(1-x2)
=log2[(x1+1)(x2+1)/(x1-1)(x2-1)]
若x=(x1+x2)/(1+x1x2)
则(1+x)/(1-x)
=[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)]
上下乘(1+x1x2)
=(1+x1x2+x1+x2)/(1+x1x2-x1-x2)
=(x1+1)(x2+1)/(x1-1)(x2-1)
所以f((x1+x2)/(1+x1x2))=log2[(x1+1)(x2+1)/(x1-1)(x2-1)]
所以f(x1)+f(x2)=f((x1+x2)/(1+x1x2))