\(A=-2x^2+8x+1\\=-2(x^2-4x)+1\\=-2(x^2-4x+4)+8+1\\=-2(x-2)^2+9\)
Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-2\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow A=-2\left(x-2\right)^2+9\le9\forall x\)
Dấu \("="\) xảy ra khi: \(x-2=0\Leftrightarrow x=2\)
Vậy \(Max_A=9\) khi \(x=2\).
\(A=-2x^2+8x-8+9=9-2\left(x^2-4x+4\right)=9-2\left(x-2\right)^2\)
Do \(\left(x-2\right)^2\ge0;\forall x\Rightarrow9-2\left(x-2\right)^2\le9\)
\(\Rightarrow A\le9;\forall x\)
Hay \(A_{max}=9\) khi \(x-2=0\Rightarrow x=2\)