\(P\left(x\right)=-x^2+13x+2012=-x^2+2.\dfrac{13}{2}x-\dfrac{169}{4}+2054\dfrac{1}{4}\)
\(=-\left(x^2-2.\dfrac{13}{2}x+\dfrac{169}{4}\right)+2054\dfrac{1}{4}=-\left(x-\dfrac{13}{2}\right)^2+2054\dfrac{1}{4}\)
Do \(-\left(x-\dfrac{13}{2}\right)^2\le0\left(\forall x\right)\Rightarrow-\left(x-\dfrac{13}{2}\right)^2+2054\dfrac{1}{4}\le2054\dfrac{1}{4}\left(\forall x\right)\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-\dfrac{13}{2}\right)^2=0\Leftrightarrow x=\dfrac{13}{2}\)
Vậy\(MaxA=254\dfrac{1}{4}\Leftrightarrow x=\dfrac{13}{2}\)
Ta có: P = \(-x^2 + 13x + 2012 \)
= \(\left(-x^2+2.x.\dfrac{13}{2}-\dfrac{169}{4}\right)+\dfrac{8217}{4}\)
= \(-\left(x^2-2.x.\dfrac{13}{2}+\dfrac{169}{4}\right)+\dfrac{8217}{4}\)
\(=-\left(x-\dfrac{13}{2}\right)^2+\dfrac{8217}{4}\) \(\leq\) \(\dfrac{8217}{4}\)
Dấu "=" xảy ra khi \(x-\dfrac{13}{2}=0\) \(\Leftrightarrow \) \(x=\dfrac{13}{2}\)
Vậy giá trị lớn nhất của P = \(\dfrac{8217}{4}\) khi x = \(\dfrac{13}{2}\)
