Question

已知函数 f（1—X/1+X)=1—X2/1+X2，则函数f﹙x﹚的解析式为 具体过程

Answer 1

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!

Answer 2

可以设y = (1-x)/(1+x),那么可以解得，x = （1-y)/(1+y)
将其代入原等式可得到
f(y) = 2y/(1+y^2)
所以：f(x) = 2x/(1+x^2)