A, $\frac{1}{2}$ - $\frac{2}{3}$ .x = $\frac{1}{4}$
⇒ $\frac{2}{3}$ .x = $\frac{1}{2}$ - $\frac{1}{4}$
⇒ $\frac{2}{3}$ .x = $\frac{1}{4}$
⇒ x = $\frac{1}{4}$ ÷ $\frac{2}{3}$
⇒ x = $\frac{3}{8}$
Vậy S= { $\frac{3}{8}$ }
B, |x-$\frac{1}{2}$| -$\frac{2}{3}$ = 0
⇒ |x-$\frac{1}{2}$| = $\frac{2}{3}$
⇒ \(\left[ \begin{array}{l}x-\frac{1}{2}=\frac{2}{3}\\x-\frac{1}{2}=\frac{-2}{3}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\frac{7}{6}\\x=\frac{-1}{6}\end{array} \right.\)
Vậy x ∈ { $\frac{7}{6}$ ; $\frac{-1}{6}$ }
