mk chỉ làm dk min thoy bạn thông cảm.
\(\frac{x^2+3x+5}{x^2+1}=\frac{2x^2+6x+10}{2\left(x^2+1\right)}=\frac{\left(x+3\right)^2+\left(x^2+1\right)}{2\left(x^2+1\right)}=\frac{\left(x+3\right)^2}{2\left(x^2+1\right)}+\frac{1}{2}\ge\frac{1}{2}\) Dấu = xảy ra khi x+3=0 <---> x=-3