Tìm \(x\), biết: \(\left| + x} \right| - =

* Hướng dẫn giải

\(\begin{array}{l}\,\left| {\dfrac{2}{3} + x} \right| - \dfrac{1}{3} = \dfrac{2}{3}\\\,\,\,\,\,\left| {\dfrac{2}{3} + x} \right|\,\,\,\,\,\,\,\,\,\, = \dfrac{2}{3} + \dfrac{1}{3}\\\,\,\,\,\,\left| {\dfrac{2}{3} + x} \right|\,\,\,\,\,\,\,\,\,\, = 1\end{array}\)

TH1:

\(\begin{array}{l}\dfrac{2}{3} + x = 1\\x = 1 - \dfrac{2}{3}\\x = \dfrac{1}{3}\,\,\,\,\end{array}\)

TH2: 

\(\begin{array}{l}\dfrac{2}{3} + x =  - 1\\x =  - 1 - \dfrac{2}{3}\\x =  - \dfrac{5}{3}\,\,\,\end{array}\)

Vậy \(x = \dfrac{1}{3}\) hoặc \(x =  - \dfrac{5}{3}\)