a) \(f\left(x\right)=\left(x-2\right)\left(x+1\right)\)
\(=x^2+x-2x-2\)
\(=x^2-x-2\)
\(=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x-2\)
\(=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{9}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)
Dấu "=" xảy ra khi \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy \(Min_{f\left(x\right)}=\dfrac{-9}{4}\) khi \(x=\dfrac{1}{2}.\)