$\frac{3}{2}$ $x$=-$\frac{5}{6}$⇒ $x$ = -$\frac{5}{9}$
|$x$+$\frac{3}{4}$| -$\frac{1}{2}$ = 0
|$x$+$\frac{3}{4}$| = $\frac{1}{2}$
⇒\(\left[ \begin{array}{l}x+ \frac{3}{4}= \frac{1}{2}\\x+ \frac{3}{4}= -\frac{1}{2}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x= -\frac{1}{4}\\x=-\frac{5}{4}\end{array} \right.\)
$Vậy$ $x$ $∈$ {-$\frac{1}{4}$ $;$ -$\frac{5}{4}$ }
$vuatoanhoc$