$x^2+x+6\\=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}\\=\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{23}{4}\\=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}$
Ta có: $\left(x+\dfrac{1}{2}\right)^2\ge 0$
$↔\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge \dfrac{23}{4}$
$→$ Dấu "=" xảy ra khi $x+\dfrac{1}{2}=0$
$↔x=-\dfrac{1}{2}$
Vậy biểu thức $x^2+x+6$ đạt GTNN là $\dfrac{23}{4}$ khi $x=-\dfrac{1}{2}$
