tìm số nguyên x thỏa mãn |x-2016|+|x-2017|+|x-2018|+|x-2019|=4

Đáp án:  `x ∈ { 2017 ; 2018 }`

Giải thích các bước giải:

Ta có: `|x – 2016| + |x – 2017| + |x – 2018| + |x – 2019|`

`= (|x – 2016| + |x – 2019|) + (|x – 2017| + |x – 2018|)`

`= (|x – 2016| + |2019 – x|) + (|x – 2017| + |2018 – x|)`

`≥ |x – 2016 + 2019 – x| + |x – 2017 + 2018 – x| = 3 + 1 = 4`

Dấu “=” xảy ra `⇔` $\left\{ \begin{array}{l}(x – 2016)(2019 – x) ≥ 0\\(x – 2017)(2018 – x) ≥ 0\end{array} \right.$

`⇔` $\left\{ \begin{array}{l}(x – 2016)(x – 2019) ≤ 0\\(x – 2017)(x – 2018) ≤ 0\end{array} \right.$

Mà $\left\{ \begin{array}{l}x – 2016 > x – 2019\\x – 2017 > x – 2018\end{array} \right.$

`⇔` $\left\{ \begin{array}{l}\left\{ \begin{array}{l}x – 2016 ≥ 0\\x – 2019 ≤ 0\end{array} \right.\\\left\{ \begin{array}{l}x – 2017 ≥ 0\\x – 2018 ≤ 0\end{array} \right.\end{array} \right.$

`⇔` $\left\{ \begin{array}{l}2016 ≤ x ≤ 2019\\2017 ≤ x ≤ 2018\end{array} \right.$

`⇔ 2017 ≤ x ≤ 2018`

Mà `x ∈ ZZ` `-> x ∈ { 2017 ; 2018 }`