Lời giải:
$\frac{1}{2}\times (\frac{2}{3}\times x+\frac{1}{2})=\frac{2}{3}\times (x-\frac{1}{3})$
$\frac{1}{2}\times \frac{2}{3}\times x+\frac{1}{4}=\frac{2}{3}\times x-\frac{2}{9}$
$\frac{1}{3}\times x+\frac{1}{4}=\frac{2}{3}\times x-\frac{2}{9}$
$\frac{1}{4}+\frac{2}{9}=\frac{2}{3}\times x-\frac{1}{3}\times x$
$\frac{17}{36}=x\times \frac{1}{3}$
$x=\frac{17}{36}: \frac{1}{3}=\frac{17}{12}$