`⇔ |x +` $\dfrac{1}{2}$ `| - 1 =` $\dfrac{2}{3}$
`⇔ |x +` $\dfrac{1}{2}$ `| =` $\dfrac{5}{3}$
Ta có `2` trường hợp:
TH1: `x +` $\dfrac{1}{2}$ `=` $\dfrac{5}{3}$
`⇒ x =` $\dfrac{7}{6}$
TH2: `x +` $\dfrac{1}{2}$ `=` -$\dfrac{5}{3}$
`⇒ x =` -$\dfrac{5}{3}$ `-` $\dfrac{1}{2}$
`=` -$\dfrac{13}{6}$
Vậy:....
