设f(x)=ax2+1/bx=c是奇函数（a,b,c属于整数）,且f(1)=2,f(2)

设f(x)=ax2+1/bx=c是奇函数（a,b,c属于整数）,且f(1)=2,f(2)
设f(x)=ax2+1/bx=c是奇函数（a,b,c属于整数）,且f(1)=2,f(2)

设f(x)=ax2+1/bx=c是奇函数（a,b,c属于整数）,且f(1)=2,f(2)
f（x)=(ax2+1)/(bx+c)是奇函数,则
f(-x)=-f(x)
f(1)=(a+1)/(b+c)=2
a+1=2(b+c)=2b+2c.(1)
f(-1)=(a+1)/(-b+c)=-f(1)=-2
a+1=-2(-b+c)=2b-2c.(2)
(1)-(2)得：2b+2c=2b-2c
c=0
a=2b-1
所以f(x)=(ax^2+1)/(bx)
f(2)=(4a+1)/(2b)
=[4(2b-1)+1]/(2b)
=[8b-3]/(2b)