\( \left| {4 - x} \right| + 2x = 3\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le 4\\ 4 - x + 2x = 3 \end{array} \right.\\ \left\{ \begin{array}{l} x > 4\\ - \left( {4 - x} \right) + 2x = 3 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \le 4\\ x = - 1\left( {tm} \right) \end{array} \right.\\ \left\{ \begin{array}{l} x > 4\\ x = \dfrac{7}{3}\left( {ktm} \right) \end{array} \right. \end{array} \right. \)
Vậy \(T = \left\{ {\dfrac{7}{3}} \right\} \)
Ta có: |4-x|+2x=3
⇔|4-x|=3-2x
\(\Leftrightarrow\left[{}\begin{matrix}4-x=3-2x\\4-x=2x-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4-x-3+2x=0\\4-x-2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}1+x=0\\7-3x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=\frac{7}{3}\end{matrix}\right.\)
Vậy: x=-1