`7 + ( 3 - 1/2 ) = | x - 5 |`
`⇔ 7 + 5/2 = | x - 5 |`
`⇔ 19/2 = | x - 5 |`
`⇔` \(\left[ \begin{array}{l}x - 5 = \frac{19}{2}\\x - 5 = \frac{- 19}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x =\frac{29}{2}\\x = \frac{- 9}{2}\end{array} \right.\)
Vậy , `x ∈ { 29/2 ; ( - 9 )/2 } .`
