(1)中令x=1,得f(2)=f(1)=0,所以a+b+c=0,4a+2b+c=0所以b=-3a,c=2af(x)=ax^2-3ax+2a=a[(x-3/2)^2-1/4]>=-1/4a对于（3）当且仅当f(x)=-1/4a时等式成立,所以-1/4a=1/4a-1/2解得a=1,b=-3,c=2所以f(x)=x^2-3x+2...