Question

已知f（1+1/x）=x/1-x²，则f（x）=

我用换元法，令1+1/x=t，x=1/t-1，再用x=1/t-1代入x/1-x²，然后就不会做了，之后过程详细点。如果我本来就错了，告诉我怎么错了，接下来该怎么做

Answer 1

f(1 + 1/x) = x/(1-x^2)

let
y = 1+ 1/x
x = 1/(y-1)

f(1 + 1/x) = x/(1-x^2)
f(y) =[1/(y-1)]/ { 1 - [1/(y-1)]^2 }
= (y-1)/(y^2-2y)
ie
f(x) = (x-1)/(x^2-2x)

追问

x≠0，f(y) =[1/(y-1)]/ { 1 - [1/(y-1)]^2 }
= (y-1)/(y^2-2y)这之间的过程呢，我就是不懂这之间的转换

追答

f(y) =[1/(y-1)]/ { 1 - [1/(y-1)]^2 }

= (y-1)/(y^2-2y)

这是化简分子，分母，然后得出的结果

1 - [1/(y-1)]^2 = [(y-1)^2 -1]/(y-1)^2
= (y^2-2y+1-1)/(y-1)^2
=(y^2-2y)/(y-1)^2

f(y) =[1/(y-1)]/ { 1 - [1/(y-1)]^2 }
=[1/(y-1)] / [(y^2-2y)/(y-1)^2]
=[1/(y-1)] . [(y-1)^2/(y^2-2y)]
= (y-1)/(y^2-2y)