f(x)=xe^x -ax^2-x >=0恒成立(x>=0)求a的取值范围

f(x)=xe^x -ax^2-x >=0恒成立(x>=0)即ax²≤xe^x-xx=0时,成立x>0时即 a≤e^x/x-1/x=(e^x-1)/x 恒成立设g(x) =(e^x-1)/x (x>0)g'(x)=(xe^x-e^x)/x²+1/x²=[e^x(x-1)+1]/x²h(x)=e^x(x-1)+1h'(x)=e^x...