a. $\frac{1}{2}$ + $\frac{3}{4}$ x = $\frac{1}{4}$
$\frac{3}{4}$ x = $\frac{1}{4}$ - $\frac{1}{2}$ = $\frac{-1}{4}$
x = $\frac{-1}{4}$ : $\frac{3}{4}$ = $\frac{-1}{3}$
Vậy x = $\frac{-1}{3}$
b. $\frac{5}{6}$ + $\frac{1}{6}$ x = -2
$\frac{1}{6}$ x = -2 - $\frac{5}{6}$ = $\frac{-17}{6}$
x = $\frac{-17}{6}$ : $\frac{1}{6}$ = -17
Vậy x = -17
c. x (x - $\frac{2}{3}$ ) = 0
⇒ \(\left[ \begin{array}{l}x=0\\x=\frac{2}{3}\end{array} \right.\)
Vậy x ∈ {0; $\frac{2}{3}$ }