\(3x-\left|2x+1\right|=2\)
\(\Rightarrow\left|2x+1\right|=3x-2\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=3x-2\\2x+1=3x+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-3x=-2-1\\2x-3x=2-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-x=-3\\-x=1\end{matrix}\right.\)
\(\Rightarrow x\in\left\{3;-1\right\}\)
\(3x-\left|2x+1\right|=2\)
th1: \(2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\ge\dfrac{-1}{2}\)
\(\Rightarrow3x-\left|2x+1\right|=2\Leftrightarrow3x-\left(2x+1\right)=2\Leftrightarrow3x-2x-1=2\)
\(\Leftrightarrow3x-2x=2+1\Leftrightarrow x=3\left(tmđk\right)\)
th2: \(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< \dfrac{-1}{2}\)
\(\Rightarrow3x-\left|2x+1\right|=2\Leftrightarrow3x-\left(-2x-1\right)=2\Leftrightarrow3x+2x+1=2\)
\(\Leftrightarrow3x+2x=2-1\Leftrightarrow5x=1\Leftrightarrow x=\dfrac{1}{5}\left(loại\right)\)
vậy \(x=3\)
