Đáp án:
$min_{x_1+x_2}=0 \Leftrightarrow x=-1\\ max_{x_1+x_2}=\dfrac{4}{3} \Leftrightarrow x=1$
Giải thích các bước giải:
$\Delta=(m^2+2m+1)^2+4(m^2+m+1)\\ =(m^2+2m+1)^2+4\left(m^2+m+\dfrac{1}{4}\dfrac{3}{4}\right)\\ =(m^2+2m+1)^2+4\left(\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right) >0 \ \forall \ m\\ Vi-et:x_1+x_2=\dfrac{m^2+2m+1}{m^2+m+1}\\ =\dfrac{(m+1)^2}{\left(\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right)} \ge \ 0 \forall \ m$
Dấu "=" xảy ra $\Leftrightarrow m=-1$
$\dfrac{m^2+2m+1}{m^2+m+1}\\ =\dfrac{\dfrac{4}{3}(m^2+m+1)-\dfrac{1}{3}(m^2-2m+1)}{m^2+m+1}\\ =\dfrac{4}{3}-\dfrac{(m-1)^2}{3(m^2+m+1)}\\ =\dfrac{4}{3}-\dfrac{(m-1)^2}{3\left(\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right) } \le \dfrac{4}{3} \ \forall \ m$
Dấu "=" xảy ra $\Leftrightarrow m=1$
