$A, |x + 3|= 3x -5$
$⇒$ \(\left[ \begin{array}{l}x+3=3x-5\\-(x+3)=3x-5\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}3+5=3x-x\\-x-3=3x-5\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}8=2x\\-x-3x=-5+3\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=4(TM)\\x=\dfrac{1}{2}(KTM)\end{array} \right.\)
Vậy $x=4$
$B, |3x| = -2x +1$
$⇒$ \(\left[ \begin{array}{l}3x=-2x+1\\-3x=-2x+1\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=-1\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}0=5(KTM)\\x=\dfrac{5}{6}(KTM)\end{array} \right.\)
Vậy $x$∈{$-1;\dfrac{1}{5}$}
