\(\left|x\right|+x=\frac{2}{3}\\ \Rightarrow\left|x\right|=\frac{2}{3}-x\left(ĐK:\frac{2}{3}-x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x=\frac{2}{3}-x\\x=x-\frac{2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{2}{3}\\0=-\frac{2}{3}\left(\text{vô lí}\right)\end{matrix}\right.\\ \Rightarrow x=\frac{2}{3}:2\\ \Rightarrow x=\frac{1}{3}\left(t/m\right)\)
Vậy \(x=\frac{1}{3}\)
\(\left|x\right|+x=\frac{2}{3}\)
⇒ \(\left|x\right|=\frac{2}{3}-x\)
⇒ \(\left[{}\begin{matrix}x=\frac{2}{3}-x\\x=x-\frac{2}{3}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x+x=\frac{2}{3}\\x-x=\frac{2}{3}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{2}{3}\\0x=\frac{2}{3}\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=\frac{2}{3}:2\\x=\frac{2}{3}:0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{1}{3}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3}\right\}.\)
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