`     | 5x - 2 | ≤ 13 `

` ⇔ -13 ≤ 5x - 2 ≤ 13 `

` ⇔ -11 ≤ 5x ≤ 15 `

` ⇔ ` $\dfrac{-11}{5 }$ ` ≤ x ≤ 3 `

 Mà ` x ∈ Z ` nên ta có : 

 ` x = { -2 ; -1 ; 0 ; 1 ; 2 ; 3 } `

 Vậy  ` x = { -2 ; -1 ; 0 ; 1 ; 2 ; 3 } ` thì  `  | 5x - 2 | ≤ 13 `

Lời giải chi tiết:

`|5x-2|\le13`

$*$`)` Xét `|5x-2|=13`

`=>`\(\left[ \begin{array}{l}5x-2=13\\5x-2=-13\end{array} \right.\)

`=>`\(\left[ \begin{array}{l}5x=15\\5x=-11\end{array} \right.\)

`=>`\(\left[ \begin{array}{l}x=3\\x=\dfrac{-11}{5}\text{(loại)}\end{array} \right.\)

$*$`)` Xét `|5x-2|<13`

`=>`\(\left[ \begin{array}{l}5x-2<13\\5x-2>-13\end{array} \right.\)

`=>`\(\left[ \begin{array}{l}5x<15\\5x>-11\end{array} \right.\)

`=>`\(\left[ \begin{array}{l}x<3\\x>\dfrac{-11}{5}\end{array} \right.\)

Suy ra`:`

`(-11)/5\lex\le3`

Mà `x∈ZZ` nên `x∈{-2;-1;0;1;2;3}`

Vậy `x∈{-2;-1;0;1;2;3}`

`@uniquetimehunter`