Đáp án:
$A_{min}=\dfrac{479}{33}$
Giải thích các bước giải:
$A=8x^2-4x+\dfrac{1}{4}x^2+15$
$=\dfrac{33}{4}x^2-4x+15$
$=\dfrac{33}{4}\bigg{(}x^2-\dfrac{16}{33}x+\dfrac{20}{11}\bigg{)}$
$=\dfrac{33}{4}\bigg{(}x^2-2.x.\dfrac{8}{33}+\dfrac{64}{1089}+\dfrac{1916}{1089}\bigg{)}$
$=\dfrac{33}{4}.\bigg{(}x-\dfrac{8}{33}\bigg{)}^2+\dfrac{479}{33}$
Ta có: $\bigg{(}x-\dfrac{8}{33}\bigg{)}^2\ge 0⇒\dfrac{33}{4}.\bigg{(}x-\dfrac{8}{33}\bigg{)}^2\ge 0$
$⇒\dfrac{33}{4}.\bigg{(}x-\dfrac{8}{33}\bigg{)}^2+\dfrac{479}{33}\ge \dfrac{479}{33}$
$⇒A\ge \dfrac{479}{33}⇒A_{min}=\dfrac{479}{33}$
Dấu "=" xảy ra khi: $\bigg{(}x-\dfrac{8}{33}\bigg{)}^2= 0$
$⇒x=\dfrac{8}{33}$
Vậy $A_{min}=\dfrac{479}{33}$ khi $x=\dfrac{8}{33}$.
