高一函数证明题f(x)=log2 (1+x)/(1-x)(1)求证：f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)](2)若f[(a+b)/(1+ab)]=1,f(-b)=1/2,求f(a)的值

(1)f(x1)+f(x2)=log2[(1+x1)/(1-x1)]+log2[(1+x2)/(1-x2)]
=log2[(1+x1+x2+x1x2)/(1-x1-x2+x1x2)]
f[(x1+x2)/(1+x1x2)]=log2{[1+(x1+x2)/(1+x1x2)]/[1-(x1+x2)/(1+x1x2)]}
=log2{[(1+x1x2+x1+x2)/(1+x1x2)]/[(1+x1x2-x1-x2)/(1+x1x2)]}
=log2[(1+x1+x2+x1x2)/(1-x1-x2+x1x2)]
所以,f(x1)+f(x2)=f[(x1+x2)/(1+x1x2)]

1）左边=f(x1)+f(x2)=log2[(x1+1)/(1-x1)]+log2[(x2+1)/(1-x2)]
=log2{[(x1+1)(x2+1)]/[(1-x1)(1-x2)]}
=log2[(x1x1+x1+x2+1)/(x1x2+1-x1-x2)]
右边=f[(x1+x2)/(1+x1x2)]=log2[(1+x1x2+x1+x2)/(1+x1x2)] / [(1+x1x2-x1-x2)/(1+x1x2)]
=log2[(x1x1+x1+x2+1)/(x1x2+1-x1-x2)]=左边

(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+x1x...