Question

已知函数﹙x－1﹚f﹙x＋1／x－1﹚＋f﹙x﹚=x,其中x≠1,求函数解析式.

Answer 1

令a=(x+1)/(x-1)=(x-1+2)/(x-1)=1+2/(x-1) 2/(x-1)=a-1 x-1=2/(a-1) x=2/(a-1)+1=(a+1)/(a-1) 代入(x-1)f[(x+1)/(x-1)]+f(x)=x [2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1) 所以[2/(x-1)]f(x)+f[(x+1)/(x-1)]=(x+1)/(x-1) ----(2) (x-1)f[(x+1)/(x-1)]+f(x)=x ---(1) (2)*(x-1)-(1) ,得 2f(x)+(x-1)f[(x+1)/(x-1)]-(x-1)f[(x+1)/(x-1)]-f(x)=(x+1)-x f(x)=1,（x≠1)