已知f((1-x)/(1+x))=(1-x^2)/(1+x^2),求f(x)解析式

来源：百度知道 编辑：UC知道 时间：2024/06/09 05:20:22

设a=(1-x)/(1+x)=-1+2/(1-x)
2/(1-x)=a+1
1-x=2/(a+1)
x=1-2/(a+1)=(a-1)/(a+1)

(1-x^2)/(1+x^2)
=1-2/(1+x^2)
=1-2/[1+(a-1)^2/(a+1)^2]
=-2a/(a^2+1)

所以f(a)=-2a/(a^2+1)
所以f(x)=-2x/(x^2+1)

令(1-x)/(1+x)=t
所以(1-x)/(1+x)=-1+2/(1+x)=t
2/(1+x)=t+1
1+x=2/(t+1)
x=(1-t)/(1+t)
所以f(t)=(1-x^2)/(1+x^2)=-1+2/(1+x^2)=-1+2/(1+(1-t)^2/(1+t)^2)=-1+(1+t)^2/(1+t^2)=2t/(1+t^2)
所以f(x)=2x/(1+x^2)

令(1-x)/(1+x)=t
1-x=t+tx
x=(1-t)/(1+t)
f(t)=(1-(1-t)^2/(1+t)^2)/(1+(1-t)^2/(1+t)^2)
=((1+t)^2-(1-t)^2)/((1+t)^2+(1-t)^2)
=(1+t+1-t)(1+t-1+t)/(2t^2+2)
=2t/(2t^2+2)
=t/(t^2+1)
把t换成x
f(x)=x/(x^2+1)