当x>1时,不等式mx^2+mx+1≥x恒成立,则实数m的取值范围是
问题描述：
当x>1时,不等式mx^2+mx+1≥x恒成立,则实数m的取值范围是
答
答：x>1时：mx^2+mx+1>=xm(x^2+x)>=x-1>0m>=(x-1)/(x^2+x)=(x-1)*[1/x-1/(x+1)]=(x-1)/x-(x-1)/(x+1)=1-1/x-(x+1-2)/(x+1)=1-1/x-1+2/(x+1)=2/(x+1)-1/x设f(x)=2/(x+1)-1/x求导：f'(x)=-2/(x+1)^2+1/x^2解f'(x)=-2/(...