Question

已知f(1+x/1-x)=1+x²/1-x²,求f(x)的解析式

Answer 1

代换法：

令(1+x)/(1-x)=t，则x=(t-1)/(t+1)
f(t)=[1+(t-1)²/(t+1)²]/[1-(t-1)²/(t+1)²]
=[(t+1)²+(t-1)²]/[(t+1)²-(t-1)²]
=(2t²+2)/(4t)
=(t²+1)/(2t)
=t/2 +1/(2t)
分式有意义，t≠0
将t换成x
f(x)的解析式为f(x)=x/2 + 1/(2x) (x≠0)

Answer 2

用换元法思路很直接

令t=(1-x)/(1+x)
(1+x)t=1-x
tx+x=1-t
x=(1-t)/(1+t)
f(t)
=(1-x²)/(1+x²)
=[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+2t+t²)-(1-2t+t²)]/[(1+2t+t²)+(1-2t+t²)]
=(4t)/(2+2t²)
=2t/(1+t²)
将t换回x，即得
f(x)=2x/(1+x²)
不懂的欢迎追问，如有帮助请采纳，谢谢!

追问

我谢谢你是f(1+x/1-x)=不是令t=(1-x)/(1+x)
复制粘贴真的大丈夫？

Answer 3