\( 3x - \left| {2x + 1} \right| = 2\\ \Leftrightarrow 3x - 2 = \left| {2x + 1} \right|\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{1}{2}\\ 3x - 2 = 2x + 1 \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{1}{2}\\ - \left( {3x - 2} \right) = 2x + 1 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{1}{2}\\ x = 3\left( {t/m} \right) \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{1}{2}\\ x = \dfrac{1}{5}\left( {ktm} \right) \end{array} \right. \end{array} \right. \)
Vậy $x=3$ là nghiệm
\(3\left(x-1\right)=2\left(y-2\right)\Rightarrow6\left(x-1\right)=4\left(y-2\right)\Rightarrow6\left(x-1\right)=4\left(y-2\right)=3\left(z-3\right)\)
\( \Rightarrow \dfrac{{6\left( {x - 1} \right)}}{{12}} = \dfrac{{4\left( {y - 2} \right)}}{{12}} = \dfrac{{3\left( {z - 3} \right)}}{{12}} \Rightarrow \dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{2} = \dfrac{{z - 3}}{4}\)
Đặt \(\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4} = k \Rightarrow x = 2k + 1;y = 3k + 2;z = 4k + 3\)
Mà \(2x+3y-z=50\Rightarrow2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)
\(\Leftrightarrow4k+2+9k+6-4k-3=50\Leftrightarrow9k=45\Leftrightarrow k=5\)
Vậy \(\left\{{}\begin{matrix}x=2.5+1=11\\y=3.5+2=17\\z=4.5+3=23\end{matrix}\right.\)
