\(3x-\left|2x+1\right|=2\)
\(\Rightarrow\left|2x+1\right|=3x-2\)
\(\Rightarrow2x+1=\pm\left(3x-2\right)\)
+) \(2x+1=3x-2\)
\(\Rightarrow1=x-2\)
\(\Rightarrow x=3\)
+) \(2x+1=-\left(3x-2\right)\)
\(\Rightarrow2x+1=-3x+2\)
\(\Rightarrow2x+3x=2-1\)
\(\Rightarrow5x=1\)
\(\Rightarrow x=\frac{1}{5}\)
Vậy \(x\in\left\{3;\frac{1}{5}\right\}\)