f(x)=e^x-1-x-ax^2,f'(x)=e^x-1-2ax当a≤1/2时,f'(x)=e^x-1≥0在x≥0时恒成立,f(x)在[0,+∞）单调递增,且f(0)=0所以当a≤1/2时,有x≥0时f(x)≥0当a＞1/2时,将f(x)=e^x-1-x-ax^2分解为h(x)=e^x-1和g(x)=x+ax^2即f(x)=...