a) $3x^2-6x+1$
$=3(x^2-2x+\dfrac{1}{3})$
$=3(x^2-2x+1-\dfrac{2}{3})$
$=3(x-1)^2-2$
Ta nhận thấy: $3(x-1)^2≥0$
$→3(x-1)^2-2≥2$
$→$ Dấu "=" xảy ra khi $x-1=0$
$→x=1$
$→A_{\min}=-2$ khi $x=1$
b) $\dfrac{3}{2}x^2+2x+3$
$=\dfrac{3}{2}(x^2+\dfrac{4}{3}x+2)$
$=\dfrac{3}{2}(x^2+2.\dfrac{2}{3}.x+\dfrac{4}{9}+\dfrac{14}{9})$
$=\dfrac{3}{2}(x+\dfrac{2}{3})^2+\dfrac{7}{3}$
Ta nhận thấy: $\dfrac{3}{2}(x+\dfrac{2}{3})^2≤0$
$→\dfrac{3}{2}(x+\dfrac{2}{3})^2+\dfrac{7}{3}≥\dfrac{7}{3}$
$→$ Dấu "=" xảy ra khi $x+\dfrac{2}{3}=0$
$→x=-\dfrac{2}{3}$
$→B_{\min}=\dfrac{7}{3}$ khi $x=-\dfrac{2}{3}$
