$(x-\dfrac{3}{6}):\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{14}{24}$
$\Leftrightarrow(x-\dfrac{3}{6}):\dfrac{1}{2}=\dfrac{14}{24}-\dfrac{1}{2}$
$\Leftrightarrow(x-\dfrac{3}{6}):\dfrac{1}{2}=\dfrac{1}{12}$
$\Leftrightarrow x-\dfrac{3}{6}=\dfrac{1}{12}×\dfrac{1}{2}$
$\Leftrightarrow x-\dfrac{3}{6}=\dfrac{1}{24}$
$\Leftrightarrow x=\dfrac{1}{24}+\dfrac{3}{6}$
$\Leftrightarrow x=\dfrac{13}{24}$
Vậy $x=\dfrac{13}{24}$
