Tìm GTNN x2 + 2021x 2x^2 – 4x +1 3x^2 – 2x

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 $x=\dfrac{1}{3}$.