Sửa: $2x^2-3x+7$
$=2(x^2-\dfrac{3}{2}x+\dfrac{7}{2})$
$=2(x^2-2.x.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{47}{16})$
$=2(x-\dfrac{3}{4})^2+\dfrac{47}{8}$
Ta nhận thấy: $2(x-\dfrac{3}{4})^2\ge 0$
$\to 2(x-\dfrac{3}{4})^2+\dfrac{47}{8}\ge \dfrac{47}{8}$
$\to$ Dấu "=" xảy ra khi $x-\dfrac{3}{4}=0$
$\to x=\dfrac{3}{4}$
$\to A_{\min}=\dfrac{47}{8}$ khi $x=\dfrac{3}{4}$
