Giải thích các bước giải:
a) $A=x^2+2021x$
$=x^2+2.x.\dfrac{2021}{2}+\dfrac{4084441}{4}-\dfrac{4084441}{4}$
$=\left(x+\dfrac{2021}{2}\right)^2-\dfrac{4084441}{4}\ge -\dfrac{4084441}{4}$
$A\ge -\dfrac{4084441}{4}⇒A_{min}=-\dfrac{4084441}{4}$
Dấu "=" xảy ra khi:
$x+\dfrac{2021}{2}=0$
$⇔x=-\dfrac{2021}{2}$
Vậy $A_{min}=-\dfrac{4084441}{4}$ khi $x=-\dfrac{2021}{2}$
b) $B=2x^2-4x+1$
$=2\left(x^2-2x+\dfrac{1}{2}\right)$
$=2\left(x^2-2x+1-\dfrac{1}{2}\right)$
$=2(x-1)^2-1\ge -1$
$B\ge -1⇒B_{min}=-1$
Dấu "=" xảy ra khi:
$x-1=0$
$⇒x=1$
Vậy $B_{min}=-1$ khi $x=1$
c) $C=3x^2-2x$
$=3\left(x^2-\dfrac{2}{3}x\right)$
$=3\left(x^2-2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}\right)$
$=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{3}\ge -\dfrac{1}{3}$
$C\ge -\dfrac{1}{3}⇒C_{min}=-\dfrac{1}{3}$
Dấu "=" xảy ra khi:
$x-\dfrac{1}{3}=0$
$⇔x=\dfrac{1}{3}$
Vậy $C_{min}=-\dfrac{1}{3}$ khi $x=\dfrac{1}{3}$.
