\(A=-2\left(4a^2-4a+1\right)+5=5-2\left(2a-1\right)^2\le5\)
\(A_{max}=5\) khi \(a=\dfrac{1}{2}\)
a) Ta có: \(A=-8a^2+8a+3\)
\(=-8\left(a^2-a-\dfrac{3}{8}\right)\)
\(=-8\left(a^2-2\cdot a\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=-8\left(a-\dfrac{1}{2}\right)^2+5\le5\forall a\)
Dấu '=' xảy ra khi \(a=\dfrac{1}{2}\)