已知函数y=f(x)满足条件f(x+1)=f(x-1),且当x€[-1,0]时,f(x)=3的x次方+4/9,则f(log以1/3为底5为真数)的值等于？

做不来

设y=x(y-1)f[(y+1)/(y-1)]+f(y)=y    (1)    设y=(x+1)/(x-1)    [(y+1)/(y-1)-1]f+f[(y+1)/(y-1)]=(y+1)/(y-1)[(y+1-y+1)/(y-1)f+f[(y+1)/(y-1)]=(y+1)/(y-1)2/(y-1)f)]}+f[(y+1)/(y-1)]=(y+1)/(y-1)2/(y-1)f(y)+f[(y+1)/(y-1)]=(y+1)/(y-1)    2f(y)+(y-1)f[(y+1)/(y-1)]=y+1    (2)(2)-(1)，得f(y)=1由y=x ，x≠-1（总感觉x≠-1这个条件不太对，应该是x≠1才对），得y≠-1由y=(x+1)/(x-1)=1+2/(x-1)，得y≠1f(y)=1   (y≠± 1)函数解析式为f(x)=1    (x≠± 1)