ta cóA = \(x^2-x+2009\)
\(\Leftrightarrow x^2-2x\dfrac{1}{2}+\dfrac{1}{4}+2008.75\)
\(\Leftrightarrow\left(x-\dfrac{1}{4}\right)^2+2008.75\)
mà \(\left(x-\dfrac{1}{4}\right)^2\ge0\Leftrightarrow\left(x-\dfrac{1}{4}\right)^2+2008.75\ge2008.75\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\dfrac{1}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{4}=0\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy Min A= 2008.75 tại \(x=\dfrac{1}{4}\)
Đặt A = x2 - x + 2009
⇔ A = x2 - \(2x.\dfrac{1}{2}\) + \(\left(\dfrac{1}{2}\right)^2\) + 2009 - \(\dfrac{1}{4}\)
⇔ A = \(\left(x-\dfrac{1}{2}\right)^2\) + 2008,75
Vì \(\left(x-\dfrac{1}{2}\right)^2\) ≥ 0 nên A ≥ 2008,75; ∀x ∈ N
Vậy Min A = 2008,75 khi \(x-\dfrac{1}{2}\) = 0
_____________________⇔ x = \(\dfrac{1}{2}\)
Chúc bạn học tốt !!!
