已知f（1+1/x）=x/1-x²,则f（x）=
问题描述：
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,如果我本来就错了,告诉我怎么错了,
答
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)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)