f（x)=(ax2+1)/(bx+c)是奇函数,则f(-x)=-f(x)f(1)=(a+1)/(b+c)=2a+1=2(b+c)=2b+2c.(1)f(-1)=(a+1)/(-b+c)=-f(1)=-2a+1=-2(-b+c)=2b-2c.(2)(1)-(2)得：2b+2c=2b-2cc=0a=2b-1所以,f(x)=(ax^2+1)/(bx)f(2)=(4a+1)/(2b)=...