f(x)=(x^2-x)(x-a)f'(x)=(x^2-x)(x-a)'+(x^2-x)'(x-a) =x^2-x+(2x-1)(x-a) =3x^2-2(a+1)x+a当x>2时f'(x)>0则应该有对称轴<=2,f'(2)>=0(这两条保证了二次函数在（2，正无穷）上是>0的即有x=(a+1)/3<=2,a<=5f(2)= 3*4-4(a+1)+a>=0解得a<=8/3
