`-3 | x + 2 | = 9`
`|x + 2 | = 9 : ( - 3 )`
`|x + 2 | = -3`
`-> x ∈ \emptyset`
`| 2x | =` $\frac{1}{3}$
=> \(\left[ \begin{array}{l}2x = \frac{1}{3} \\2x = -\frac{1}{3}\end{array} \right.\)
`TH1 : 2x =` $\frac{1}{3}$
`x =` $\frac{1}{3}$ : `2`
`x` = $\frac{2}{3}$
`5 | x - 1 | = 10`
`| x - 1 | = 10 : 5 = 2`
\(\left[ \begin{array}{l}x - 1 = 2\\x - 1 = - 2\end{array} \right.\)
`TH1 : x - 1 = 2`
`x = 2 + 1 = 3`
`TH2 : x - 1 = - 2`
`x = -2 + 1 = -1`
Vậy `x ∈ { 3, -1 }`
