Đáp án:
$max_A=\dfrac{11}{2} \Leftrightarrow x=\dfrac{3}{2}.$
Giải thích các bước giải:
$A=-2x^2+6x+1\\ =-2\left(x^2-3x-2\right)\\ =-2\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{11}{4}\right)\\ =-2\left(\left(x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\right)\\ =-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2} \le \dfrac{11}{2} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{3}{2}=0 \Leftrightarrow x=\dfrac{3}{2}$
Vậy $max_A=\dfrac{11}{2} \Leftrightarrow x=\dfrac{3}{2}.$
