f(-x)=-f(x)(ax²+1)/(-bx+c)=-(ax²+1)/(bx+c)所以-bx+c=-bx-cc=0f(1)=(a+1)/b=2a=2b-1f(2)=(4a+1)/2b<3(8b-3)/2b-3=(2b-3)/2b<0所以b(2b-3)<00b=1所以f(x)=(x²+1)/x
