b: Ta có: \(x^2-x+5\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{19}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\forall x\)
\(\Leftrightarrow\dfrac{2022}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}}\le\dfrac{8088}{19}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)