\(|3x+1|=2x-1\)
\(\Rightarrow|3x+1|-2x=-1\)
\(\Rightarrow\left[{}\begin{matrix}3x+1-2x=-1,3x+1\ge0\\-\left(3x+1\right)-2x=-1,3x+1< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2,x\ge-\frac{1}{3}\\x=0,x< -\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)
ĐK: \(x\ge\frac{1}{2}\)
\(\left|3x+1\right|=2x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=2x-1\\3x+1=-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)( không thỏa mãn )
Vậy \(x\in\varnothing\)
