$a) A=16x^2+8x+5$
$=16x^2+8x+1+4$
$=(4x+1)^2+4$
Vì $(4x+1)^2 \geq 0 ⇒ (4x+1)^2+4 \geq 0$
$⇒ A \geq 4$
Dấu "=" xảy ra khi $4x+1=0 ⇔ x=\dfrac{-1}{4}$
Vậy $A_{Min}=4$ khi $x=\dfrac{-1}{4}$
$b) B=2x^2-5x$
$=2x^2-5x+\dfrac{25}{8}-\dfrac{25}{8}$
$=2\left (x^2-\dfrac{5}{2}x+\dfrac{25}{16} \right )-\dfrac{25}{8}$
$=2\left (x-\dfrac{5}{4} \right )^2-\dfrac{25}{8}$
$2\left (x-\dfrac{5}{4} \right )^2 \geq 0 ⇒ 2\left (x-\dfrac{5}{4} \right )^2-\dfrac{25}{8} \geq -\dfrac{25}{8}$
$⇒B \geq -\dfrac{25}{8}$
Dấu "=" xảy ra khi $x-\dfrac{5}{4}=0 ⇔ x=\dfrac{5}{4}$
Vậy $B_{Min}=-\dfrac{25}{8}$ khi $x=\dfrac{5}{4}$
