f(x)=(lnx+m)/e^x=(lnx+m)e^(-x)f'(x)=[e^(-x)]/x-e^(-x)(lnx+m)g(x)=x+lnx+e^xf'(x)=x+lnx+1/x-(lnx+m)g'(x)=1+1/x -1/x² -1/x=1-1/x²令g'(x)=0,得x=1或x=-1(舍去)g(x)的最小值是1,所以g(1)=1 即g(1)=1+0...