设f(x)=ax^2+bx+cf(x+1)+f(x-1)=a(x+1)^2+b(x+1)+c+a(x-1)^2+b(x-1)+c=ax^2+2ax+a+bx+b+c+ax^2-2ax+a+bx-b+c=2ax^2+2bx+2a+2c=2x^2-4x对应项系数相等2a=2,2b=-4,2a+2c=0a=1,b=-2,c=-a=-1f(x)=x^2-2x-1