f[(x+1)/(3x-1)]=2f(x)+x 求 f(x) 的解析式
问题描述：
f[(x+1)/(3x-1)]=2f(x)+x 求 f(x) 的解析式
答
令a=(x+1)/(3x-1)
3ax-a=x+1
(3a-1)x=a+1
x=(a+1)/(3a-1)
所以f(a)=2f[(a+1)/(3a-1)]+(a+1)/(3a-1)
即f(x)=2f[(x+1)/(3x-1)]+(x+1)/(3x-1)
所以f[(x+1)/(3x-1)]=[f(x)-(x+1)/(3x-1)]/2
所以[f(x)-(x+1)/(3x-1)]/2=2f(x)+x
f(x)-(x+1)/(3x-1)=4f(x)+2x
3f(x)=-(x+1)/(3x-1)-2x=(-x-1-6x^2+2x)/(3x-1)=(-6x^2+x-1)/(3x-1)
f(x)=(-6x^2+x-1)/(9x-3)