Lời giải:
Ta có:
\(x^2-x+2018=x^2-2.x.\frac{1}{2}+(\frac{1}{2})^2+\frac{8071}{4}\)
\(=(x-\frac{1}{2})^2+\frac{8071}{4}\)
Ta thấy rằng \((x-\frac{1}{2})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow x^2-x+2018=(x-\frac{1}{2})^2+\frac{8071}{4}\geq \frac{8071}{4}\)
Do đó \((x^2-x+2018)_{\min}=\frac{8071}{4}\)
Dấu bằng xảy ra khi \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)