a)\(A=5-8x-2x^2\)
\(=-2\left(x^2+4x-\frac{5}{2}\right)\)
\(=-2\left(x^2+4x+4-\frac{13}{2}\right)\)
\(=-2\left[\left(x+2\right)^2-\frac{13}{2}\right]\)
\(=-2\left[\left(x+2\right)^2\right]+13\le13\)
Vậy \(A_{max}=13\Leftrightarrow x+2=0\Leftrightarrow x=-2\)