先求x>=0时函数式:设x>=0时 f(x)=ax^2+bx+c,因f(0)=0,故c=0由F(x+1)=f(x)+x+1得f(x+1)=a(x+1)^2+b(x+1)=ax^2+(2a+b)x+a+bf(x)+x+1=ax^2+bx +x+1 ax^2+(2a+b)x+a+b=ax^2+(b+1)x +1比较系数得,a+b=1,b+1=2a+b求得a=b= ...