\(A=\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{21}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\)
Vậy \(min_A=\dfrac{-21}{4}\) đạt tại \(x=\dfrac{5}{2}\)
\(A=x^2-5x+1=x^2+5x-\dfrac{21}{4}+\dfrac{25}{4}=\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{21}{4}\)
\(=\left[x^2+2.\dfrac{5}{2}.x+\dfrac{5}{2}^2\right]-\dfrac{21}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\)
Do \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\forall x\)
Dấu '' = '' xảy ra khi \(x=-\dfrac{5}{2}\)
Vậy \(min_A=-\dfrac{21}{4}\) khi \(x=-\dfrac{5}{2}\)
Ta có:
\(A=x^2-5x+1\)
\(=x^2-5x+\left(\dfrac{25}{4}-\dfrac{21}{4}\right)\)
\(=x^2-5x+\dfrac{25}{4}-\dfrac{21}{4}\)
\(=x^2-2.\dfrac{5}{2}.x+\left(\dfrac{5}{2}\right)^2-\dfrac{21}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge\dfrac{-21}{4}\forall x\)
\(\Rightarrow Min_A=\dfrac{-21}{4}\)
Dấu "=" xảy ra khi \(\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(Min_A=\dfrac{-21}{4}\) khi \(x=\dfrac{5}{2}\)
*HĐT số 2: \(\left(a-b\right)^2=a^2-2ab+b^2\)
