f(1/x)=1+x2/1-x2 则f（x）=x2+1/x2-1

f(1/x)=1+x2/1-x2 则f（x）=x2+1/x2-1
f(1/x)=1+x2/1-x2 则f（x）=x2+1/x2-1

f(1/x)=1+x2/1-x2 则f（x）=x2+1/x2-1
我来解等等
令1/x=t,所以x=1/t
f（t）=1+(1/t^2)/[1-(1/t^2)]
=1+1/(t^2-1)
=(t^2-1+1)/(t^2-1)
=(t^2)/(t^2-1)
换回即f（x）=x2/（x2-1）

F（1\x）=1+x^2\(1-x^2) 令X=1\x 则F(X)=1+（1\X^)\(1-1\(X^))=1+1\(X^-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)
...

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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)

收起

由f(1/x)=(1+x2)/(1-x2)，带入x=1/a，得f(1/(1/a))=f(a)=(a2+1)/(a2-1),从而f(x)=(x2+1)/(x2-1).希望对你有帮助，祝学习进步！