设(1-x)/(1+x)=y 则 1-x=y+xy (1+y)x=1-y 即 x=(1-y)/(1+y)∴f(y)=[1-(1-y)^2/(1+y)^2]/[1+(1-y)^2/(1+y)^2]f(y)=[(1+y)^2-(1-y)^2]/[(1+y)^2+(1-y)^2]f(y)=4y/(2+2y^2)=2y/(1+y^2)即 f(x)=2x/(1+x^2)
