$A=-x^2+3x+1$
$=-x^2+3x-\dfrac94+\dfrac{13}{4}$
$=-\left(x^2-3x+\dfrac94\right)+\dfrac{13}{4}$
$=-\left(x-\dfrac32\right)^2+\dfrac{13}{4} \le \dfrac{13}{4}$
Đẳng thức xảy ra $\Leftrightarrow x-\dfrac32=0\Leftrightarrow x=\dfrac32$
Vậy $\min A=\dfrac{13}{4}$ đạt được khi $x=\dfrac32$