f(x-1分之x+1）=3f（x）-2x，求f（x）

设 y=(x+1)/(x-1)=1+2/(x-1)，(y-1)(x-1)=2，x-1=2/(y-1)，x=1+2/(y-1)=(y+1)/(y-1)。∴ f(x)=f[(y+1)/(y-1)]。.∵ f[(x+1)/(x-1)]=3f(x)-2x∴ f(x)=3f[(x+1)/(x-1)]-2(x+1)/(x-1)=3[3f(x)-2x]-2(x+1)/(x-1)=9f(x)-6x-2(x+1)/(x-1)∴ 9f(x)-f(x)=6x+2(x+1)/(x-1)=2[3x(x-1)+(x+1)]/(x-1)∴ f(x)=(3x²-2x+1)/[4(x-1)]

令t=(x+1)/(x-1)→x=(t+1)/(t-1)f[(t+1)/(t-1)]=3f（t）-2t=3f[(x+1)/(x-1)]-2(x+1)/(x-1)f(x)=3f[(x+1)/(x-1)]-2(x+1)/(x-1)y1=f(x),y2=f[(x+1)/(x-1)]y1=3y2--2(x+1)/(x-1)y2=3y1-2xy1=3*(3y1-2x)-2(x+1)/(x-1)以下略