Mình làm cả hai cách; mình nghĩ cách hai dễ hiểu hơn @@
$C1$. $A = |x-2021| + |x-2020|$
Nếu $x ≥ 2021$ $⇒$ $|x-2021| = x - 2021$
$|x-2020| = x-2020$
$⇒ A = x- 2021 + x - 2020 = 2x - 4041$
$⇒ A ≥ 2 . 2021 - 4041 = 4042 - 4041 =1$
Nếu $2020 ≤ x < 2021$ $⇒$ $|x-2021| = -x + 2021$
$|x-2020| = x - 2020$
$⇒$ $A = -x + 2021 + x - 2020 = 1$
Nếu $x$ $<$ $2020$ $⇒$ $|x-2021| = -x + 2021$
$|x-2020| = -x + 2020$
$⇒ A = -x + 2021 - x + 2020 = -2x + 4041$
$⇒ A > -2 . 2021 + 4041 = -1 $
Vì : $|x-2021| + |x-2020|$ $≥$ $0$ $∀$ $x$
$⇒ A ≥ 1$
$⇒$ $A$ đạt $GTNN=1$ khi $2020 ≤ x < 2021$
$C2$. $A =|x-2021| + |x-2020|= |x-2021| + |2020-x|$
Áp dụng : $|a| + |b|$ $≥$ $|a+b|$ dấu "$=$" khi $a.b ≥ 0$
$⇒ A ≥ | x-2021+2020-x| = |-1| = 1$
Dấu "$=$" khi $(x-2021).(2020-x)$ $≥$ $0$ hay $x-2021;2020-x$ cùng dấu
$TH1$. $\left\{\begin{matrix}x-2021 > 0 & \\ 2020 - x > 0 & \end{matrix}\right.$ $⇒$ $KTM$ vì $x>2021$ mà lại bé hơn $2020$
$TH2$. $\left\{\begin{matrix}x-2021 < 0 & \\ 2020 - x < 0 & \end{matrix}\right.$ $⇒$ $2020 < x < 2021$ ($TM$)
$TH3$. $\left\{\begin{matrix}x-2021 = 0 ⇔ x = 2021 & \\ 2020 - x = 0 ⇔ x = 2020 & \end{matrix}\right.$
Vậy $A$ đạt $GTNN=1$ khi $2020 ≤ x ≤ 2021$.
