`2 . | x - 5 | + 3 . | 5 - x | = 10`
`⇔ 2 . | x - 5 | + 3 . | x - 5 | = 10`
`⇔ | x - 5 | . ( 2 + 3 ) = 10`
`⇔ | x - 5 | . 5 = 10`
`⇔ | x - 5 | = 10 : 5`
`⇔ | x - 5 | = 2`
`⇔` \(\left[ \begin{array}{l}x - 5 = 2\\x - 5 = - 2\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = 7\\x = 3\end{array} \right.\)
Vậy `, x ∈ { 7 ; 3 } .`