Đặt \(B=-x^2+13x+1\)
\(\Rightarrow B=-x^2+13x-\dfrac{169}{4}+\dfrac{173}{4}\)
\(\Rightarrow B=-\left(x^2-13x+\dfrac{169}{4}\right)+\dfrac{173}{4}\)
\(\Rightarrow B=-\left[x^2-2.x.\dfrac{13}{2}+\left(\dfrac{13}{2}\right)^2\right]+\dfrac{173}{4}\)
Do :\(B=-\left(x-\dfrac{13}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+\dfrac{13}{2}\right)^2+\dfrac{173}{4}\le\dfrac{173}{4}\forall x\)
\(\Rightarrow B\le\dfrac{173}{4}\)
\(\Rightarrow B_{max}=\dfrac{173}{4}\)
Dấu '=' xảy ra khi và chỉ khi \(x-\dfrac{13}{2}=0\Rightarrow x=\dfrac{13}{2}\)
Vậy \(B_{max}=\dfrac{173}{4}\) khi \(x=\dfrac{13}{2}\)
\(-x^2+13x+1\)
\(=-\left(x^2-13x-1\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{13}{2}+\dfrac{169}{4}-\dfrac{173}{4}\right)\)
\(=-\left(x-\dfrac{13}{2}\right)^2+\dfrac{173}{4}< =\dfrac{173}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{13}{2}=0\)
=>\(x=\dfrac{13}{2}\)
