设f(x)=ax^2+bx+cf(x+1)=ax^2+2ax+a+bx+b+c=ax^2+(2a+b)x+a+b+cf(x-1)=ax^2-2ax+a+bx-b+c=ax^2+(-2a+b)x+a-b+cf(x+1)＋f(x+1)=2ax^2+2bx+2a+2c对照系数2a=22b=32a+2c=5解得a=1 b=3/2 c=3/2f(x)=x^2+3/2x+3/2