\(F=x^2-x+1\)
\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\)
Vậy \(Min_F=\dfrac{3}{4}khi\)\(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)