\(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3+2x=10\left(đk:x-3\ge0\right)\\-\left(x-3\right)+2x=10\left(đk:x-3< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=10\\-x+3+2x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=10+3\\x+3=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=13\\x=10-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{3}\left(đk:x\ge3\right)\\x=7\left(đk:x< 3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{3}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\dfrac{13}{3}\)
