\(\left|x\right|+x=\frac{1}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{1}{3}-x\)
+) Xét \(x\ge0\)
\(pt\Leftrightarrow x=\frac{1}{3}-x\)
\(\Leftrightarrow2x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)( thỏa )
+) Xét \(x< 0\)
\(pt\Leftrightarrow x=x-\frac{1}{3}\)
\(\Leftrightarrow0x=\frac{-1}{3}\)( vô lí )
Vậy....
\(\left|x\right|+x=\frac{1}{3}\)
\(\Rightarrow\left|x\right|=\frac{1}{3}-x\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}-x\\-x=\frac{1}{3}-x\end{matrix}\right.\)
\(\Rightarrow2x=\frac{1}{3}\)
\(\Rightarrow\)x=\(\frac{1}{6}\)
Vậy x=\(\frac{1}{6}\)