Đáp án:

$x\in\{2017;2018\}$

Giải thích các bước giải:

$\begin{array}{l} \quad |x-22016| + |x-2017| + |x-2018| + |x-2019| =4\qquad (*)\\ \text{Ta có:}\\ +) \quad |x-2016| + |x-2018|\\ = |x-2016| + |2018-x|\\ \geq |x-2016 + 2018 - x| = 2\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow (x-2016)(2018-x) \geq 0\\ \Leftrightarrow 2016 \leq x \leq 2018\\ +) \quad |x-2017| + |x-2019|\\ = |x-2017| + |2019- x|\\ \geq |x-2017+ 2019 -x| = 2\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow (x-2017)(2017-x) \geq 0\\ \Leftrightarrow 2017 \leq x \leq 2019\\ \text{Do đó:}\\ |x-22016| + |x-2017| + |x-2018| + |x-2019| \geq 4\\ \text{Ta được:}\\ (*) \Leftrightarrow \begin{cases}|x-2016| + |x-2018| =2\\|x-2017| + |x-2019|=2 \end{cases}\\ \Leftrightarrow \begin{cases}2016 \leq x \leq 2018\\2017 \leq x \leq 2019\end{cases}\\ \Leftrightarrow 2017 \leq x \leq 2018\\mà\,\,x \in \Bbb Z\\nên\,\,x \in \{2017;2018\} \end{array}$

Đáp án:

|x−2016|+|x−2019|=|x−2016|+|2019−x||x−2016|+|x−2019|=|x−2016|+|2019−x|

≥|(x−2016)+(2019−x)|=|3|=3(1)≥|(x−2016)+(2019−x)|=|3|=3(1)

Dấu ′="⇔(x−2016)(2019−x)≥0⇔2016≤x≤2019′="⇔(x−2016)(2019−x)≥0⇔2016≤x≤2019

|x−2017|+|x−2018|=|x−2017|+|2018−x||x−2017|+|x−2018|=|x−2017|+|2018−x|

≥|(x−2017)+(2018−x)|=|1|=1(2)≥|(x−2017)+(2018−x)|=|1|=1(2)

Dấu ′="⇔(x−2016)(2019−x)≥0⇔2017≤x≤2018′="⇔(x−2016)(2019−x)≥0⇔2017≤x≤2018

(1)+(2):|x−2016|+|x−2017|+|x−2018|+|x−2019|≥4(1)+(2):|x−2016|+|x−2017|+|x−2018|+|x−2019|≥4

Để :|x−2016|+|x−2017|+|x−2018|+|x−2019|=4:|x−2016|+|x−2017|+|x−2018|+|x−2019|=4

⇒|x−2016|+|x−2019|=3⇔2016≤x≤2019(x∈Z)⇒|x−2016|+|x−2019|=3⇔2016≤x≤2019(x∈Z)

⇒x=2016;2017;2018;2019(3)⇒x=2016;2017;2018;2019(3)

|x−2017|+|x−2018|=1⇔2017≤x≤2018|x−2017|+|x−2018|=1⇔2017≤x≤2018

⇒x=2017;2018(4)⇒x=2017;2018(4)

Giải thích các bước giải: