Đáp án đúng: B
Giải chi tiết:\(\begin{array}{l}\,\left| {\frac{2}{3} + x} \right| - \frac{1}{3} = \frac{2}{3}\\\,\,\,\,\,\left| {\frac{2}{3} + x} \right|\,\,\,\,\,\,\,\,\,\, = \frac{2}{3} + \frac{1}{3}\\\,\,\,\,\,\left| {\frac{2}{3} + x} \right|\,\,\,\,\,\,\,\,\,\, = 1\end{array}\)
TH1: \(\frac{2}{3} + x \ge 0\) hay \(x \ge \frac{{ - 2}}{3}\) ta có:
\(\begin{array}{l}\left| {\frac{2}{3} + x} \right| = 1\\ \Rightarrow \frac{2}{3} + x = 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 - \frac{2}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{1}{3}\,\,\,\,\left( {TM} \right)\end{array}\)
TH2: \(\frac{2}{3} + x < 0\) hay \(x < \frac{{ - 2}}{3}\) ta có:
\(\begin{array}{l}\left| {\frac{2}{3} + x} \right| = 1\, \Rightarrow \frac{2}{3} + x = - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = - 1 - \frac{2}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = - \frac{5}{3}\,\,\,\left( {TM} \right)\end{array}\)
Vậy \(x = \frac{1}{3}\) hoặc \(x = - \frac{5}{3}\)
Chọn B
