Ta có:
\(A=\left|x-2020\right|+\left|x-2021\right|\)
\(=\left|x-2020\right|+\left|2021-x\right|\)
\(\ge\left|x-2020+2021-x\right|=\left|1\right|=1\)
Dấu "=" xảy ra khi: \(\left(x-2020\right)\left(2021-x\right)\ge0\)
\(\Rightarrow2020\le x\le2021\)
Vậy Min(A) = 1 khi \(2020\le x\le2021\)
Ta có A = |x - 2020| + |x - 2021|
= |x - 2020| + |2021 - x|
\(\ge\)|x - 2020 + 2021 - x| = |1| = 1
Dấu "=" xảy ra <=> \(\left(x-2020\right)\left(2021-x\right)\ge0\)
Xét các trường hợp
TH1 : \(\hept{\begin{cases}x-2020\ge0\\2021-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2020\\x\le2021\end{cases}}\Rightarrow2020\le x\le2021\)
TH2 : \(\hept{\begin{cases}x-2020\le0\\2021-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2020\\x\ge2021\end{cases}}\left(\text{loại}\right)\)
Vậy Min A = 1 <=> \(2020\le x\le2021\)
A = | x - 2020 | + | x - 2021 |
= | x - 2020 | + | -( x - 2021 ) |
= | x - 2020 | + | 2021 - x | ≥ | x - 2020 + 2021 - x | = | 1 | = 1
Dấu "=" xảy ra khi ab ≥ 0
=> ( x - 2020 )( 2021 - x ) ≥ 0
=> 2020 ≤ x ≤ 2021
=> MinA = 1 <=> 2020 ≤ x ≤ 2021
là:
