$x^2-x+1\\=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\\=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}$
Ta có: $\left(x-\dfrac{1}{2}\right)^2\ge 0$
$↔\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge \dfrac{3}{4}$
$→$ Dấu "=" xảy ra khi $x-\dfrac{1}{2}=0$
$↔x=\dfrac{1}{2}$
Vậy $x^2-x+1$ đạt GTNN là $\dfrac{3}{4}$ khi $x=\dfrac{1}{2}$
