\(A=-5x^2-4x+1\)
\(=-5x^2-2\sqrt{5}x.\dfrac{4}{2\sqrt{5}}-\dfrac{4}{5}+\dfrac{4}{5}+1\)
\(=-\left(\sqrt{5}x+\dfrac{4}{2\sqrt{5}}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)
\(A_{max}=\dfrac{9}{5}khi\sqrt{5}x+\dfrac{4}{2\sqrt{5}}=0\Leftrightarrow x=-\dfrac{2}{5}\)
\(A=-5x^2-4x-\dfrac{4}{5}+\dfrac{9}{5}\\ A=-\left(5x^2+4x+\dfrac{4}{5}\right)+\dfrac{9}{5}\\ A=-5\left(x^2+\dfrac{4}{5}x+\dfrac{4}{25}\right)+\dfrac{9}{5}\\ A=-5\left[x^2+2\cdot x\cdot\dfrac{2}{5}+\left(\dfrac{2}{5}\right)^2\right]+\dfrac{9}{5}\\ A=-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\\ Do\left(x+\dfrac{2}{5}\right)^2\ge0\forall x\\ \Rightarrow-\left(x+\dfrac{2}{5}\right)^2\le0\forall x\\ \Rightarrow-5\left(x+\dfrac{2}{5}\right)^2\le0\forall x\\ \Rightarrow A=-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x+\dfrac{2}{5}\right)^2=0\\ \Leftrightarrow x+\dfrac{2}{5}=0\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(A_{\left(Max\right)}=\dfrac{9}{5}\) khi \(x=-\dfrac{2}{5}\)
