b. \(⇔ \left[ \begin{array}{l}2x-3=5\\2x-3=-5\end{array} \right.\\ ⇔ \left[ \begin{array}{l}2x=8\\2x=-2\end{array} \right.\\ ⇔ \left[ \begin{array}{l}x=4\\x=-1\end{array} \right. \)
Vậy $x=\{4;-1\}$
b. \(⇔ \left[ \begin{array}{l}2x-1=2x+3\\2x-1=-2x-3\end{array} \right.\\ ⇔ \left[ \begin{array}{l}0x=4\ (Loại) \\4x=-2\end{array} \right.\\ ⇔ x=-\dfrac12 \)
Vậy $x=-\dfrac12$