f(x)+f[(x-1)/x]=1+x ==》f(x)=1+x-f[(x-1)/x],(1)(移項得)
==》f[(x-1)/x]=1+(x-1)/x-f{[(x-1)/x-1]/[(x-1)/x]}=1+(x-1)/x-f[-1/(x-1)],(2)(將(1)式中的x換成(x-1)/x得)
f[-1/(x-1)]=1+[-1/(x-1)]-f{[-1/(x-1)-1]/[-1/(x-1)]}=1-1/(x-1)-f(x),(3)(將(1)式中的x換成-1/(x-1)得)
綜合(1)(2)(3)得
f[(x-1)/x]=1+(x-1)/x-f[-1/(x-1)]=1+(x-1)/x-[1-1/(x-1)-f(x)]=(x-1)/x+1/(x-1)+f(x),
f(x)=1+x-f[(x-1)/x]=1+x-[(x-1)/x+1/(x-1)+f(x)]=1+x-(x-1)/x-1/(x-1)-f(x)=x+1/x+1/(1-x)-f(x),
2f(x)=x+1/x+1/(1-x),
f(x)=x/2+1/(2x)+1/[2(1-x)],