Question

已知函数f(x)满足(x-1)f(x-1分之x+1)+f(x)=x（x≠1),求fx的解析式

Answer 1

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

用(x+1)/(x-1)替代x

[(x+1)/(x-1)-1]f([(x+1)/(x-1)+1]/[(x+1)/(x-1)-1])+f((x+1)/(x-1))=(x+1)/(x-1)

(1) [2/(x-1)]f(x)+f((x+1)/(x-1))=(x+1)/(x-1)

联立

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

(1)*(x-1)-(2)

f(x)=1