构造函数令g(x)=f(x)/e^x求导得：g′(x)=[f′(x)e^x-f(x)e^x]/(e^x)²=[f′(x)-f(x)]/(e^x)因x取一切实数时有满足f'(x)＞f(x)故g′(x)>0即g(x)在R上单调递增故有g(a)>g(0)即f(a)/e^a>f(0)/e^0=f(0)即是f(a)>e^af...