设f(x)=axx+bx+c (a,b为常数）f(0)=c=3所以f(x)=axx+bx+3f(x+2)=a(x+2)(x+2)+b(x+2)+3=axx+(4a+b)x+(2b+4a+3)f(x+2)-f(x)=4ax+2b+4a=4x+2所以 4a=4 , 2b+4a=2 解方程组的a=1,b=-1所以 f(x)=xx-x+3