$\dfrac{3}{2x^2+2x+3}$
$=\dfrac{3}{2(x^2+x+\dfrac{3}{2})}$
$=\dfrac{3}{2(x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{5}{4})}$
$=\dfrac{3}{2(x+\dfrac{1}{2})^2+\dfrac{5}{2}}$
Ta thấy: $2(x+\dfrac{1}{2})^2+\dfrac{5}{2}\ge \dfrac{5}{2}$
$\to$ Dấu "=" xảy ra khi $x+\dfrac{1}{2}=0$
$\to x=-\dfrac{1}{2}$
$\to \max{A}=\dfrac{6}{5}$ khi $x=-\dfrac{1}{2}$