Question

已知f【(1+x)除以(1-x)】=1+x平方除以1－x平方，求f(x)

Answer 1

令x=tan u;
则(1+x)除以(1-x)=(1+tan u)除以(1-tan u)=(tan π/4 + tan u)除以(1-tan π/4 · tan u)=tan(π/4+u).
而 1+x平方除以1－x平方=(1+tan^2 u)/(1-tan^2 u)=1/cos 2u
=1/sin(π/2+2u)=[1+ tan^2 (π/4+u)]除以[2·tan(π/4+u)]
令X=tan(π/4+u),
则f(X)=(1+X平方)除以(2X)

Answer 2

已知f[(1+x)/(1-x)]=(1+x)^2/(1-x)^2=[(1+x)/(1-x)]^2
设(1+x)/(1-x)=u
则f(u)=u^2
∴f(x)=x^2 (x≠1)

Answer 3

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