*Lời giải :
Đặt `A = |-x + 2020| + |x - 1|`
Áp dụng BĐT : `|a| + |b| ≥ |a + b|`
`⇔ |-x + 2020| + |x - 1| ≥ |-x + 2020 + x -1 | = |2019| = 2019`
`-> A_{min} = 2019`
Khi và chỉ khi :
`(-x + 2020) (x - 1) ≥ 0`
`⇔` \(\left\{ \begin{array}{l}-x+2020≥0\\x-1≤0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x≥2020\\x≤1\end{array} \right.\) `⇔ 2020 ≤ x ≤ 1` (Vô lí)
`⇔` \(\left\{ \begin{array}{l}-x+2020≤0\\x-1≥0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x≤2020\\x≥1\end{array} \right.\) `⇔ 1 ≤ x ≤ 2020` (Thỏa mãn)
Vậy `A_{min} = 2019` tai `1 ≤ x ≤ 2020`
