f(x)=e^x-1-x-axf'(x)=e^x-(a+1)若a+1≤0,也即a≤-1,则f'(x)>0,f(x)严格单增,故只需f(0)≥0,1-1-(a+1)*0≥0,得0≥0恒成立.故a≤-1时满足题意.若a+1>0,也即a>-1,则方程f'(x)=e^x-(a+1)=0有实数解x=ln(a+1).此时f''(x)...