Ta có:
\(\left|x+x\right|=\frac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+x\right|=\frac{1}{3}\\\left|x+x\right|=-\frac{1}{3}\end{matrix}\right.\)
Vì \(\left|x+x\right|\ge0\) mà \(-\frac{1}{3}< 0\Rightarrow\) TH2 bị loại.
Lại có:
\(\left|x+x\right|=\frac{1}{3}\)
\(\Rightarrow\left|x+x\right|=\frac{1}{6}+\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{6}\)
\( \left| x \right| + x = \dfrac{1}{3}\\ \Leftrightarrow \left[ \begin{array}{l} x + x = \dfrac{1}{3}\left( {x \ge 0} \right)\\ - x + x = \dfrac{1}{3}\left( {x < 0} \right) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = \dfrac{1}{3}\\ 0x = \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{1}{6} \text{(thỏa mãn điều kiện)}\\ 0x = \dfrac{1}{3}\text{(vô nghiệm)} \end{array} \right. \)
Vậy \(x = \dfrac{1}{6} \)
