已知f(x)=lg1+x分之1-x,且f(x)+f(y)=f(z),则z=?

Question

已知f(x)=lg1+x分之1-x,且f(x)+f(y)=f(z),则z=?

Answer 1

f(x)=lg[(1-x)/(1+x)]
f(y)=lg[(1-y)/(1+y)]
f(x)+f(y)=lg[(1-x)/(1+x)]+lg[(1-y)/(1+y)]
=lg[(1-x)(1-y)/(1+x)(1+y)]
=lg[(1-x-y+xy)/(1+x+y+xy)]
=lg{[(1+xy)-(x+y)]/[(1+xy)+(x+y)]} (同除以1+xy)
=lg{[1-(x+y)/(1+xy)]/[1+(x+y)/(1+xy)]}
=f(z)
=lg[(1-z)/(1+z)]
对比,得z=(x+y)/(1+xy)