对x求导,得f'(x)=[e^(x-m)] -1当m＞1时,f'(0)=e^(-m) -1＜0f'(m)=[e^0] -1=0也就是在区间(0,m)内f'(x)恒小于0,即f(x)在此区间递减f(0)=[e^(-m)] -0＞0f(m)=[e^0] -m=1-m＜0∴在(0,m)上必存在一点n使得,f(n)=0即在此...