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

Answer 2

有问题吧，怎么解出来f(x)=1呀
哪里有问题吧

Answer 3

(x-1)f[(x+1)/(x-1)]+f(x)=x (1)
令(x+1)/(x-1)=t，则x=(t+1)/(t-1)
[(t+1)/(t-1) -1]f(t) +f[(t+1)/(t-1)]=(t+1)/(t-1)
f[(t+1)/(t-1)]+2f(t)=(t+1)/(t-1)
t同样在定义域上取值，将t换成x
f[(x+1)/(x-1)]+2f(x)=(x+1)/(x-1) (2)
(2)×(x-1) -(1)
[2(x-1) -1]f(x)=(x+1)-x
(2x-3)f(x)=1
f(x)=1/(2x-3)

(x+1)/(x-1)有意义，x≠1；1/(2x-3)有意义，2x-3≠0 x≠3/2

函数解析式为f(x)=1/(2x-3) (x≠1且x≠3/2)