\(\dfrac{1}{5}\times x-\dfrac{2}{3}=\dfrac{1}{10}\times x+\dfrac{5}{6}\)
\(\dfrac{1}{5}x-\dfrac{2}{3}-\dfrac{1}{10}x-\dfrac{5}{6}=0\)
\(\dfrac{1}{5}x-\dfrac{1}{10}x-\dfrac{2}{3}-\dfrac{5}{6}=0\)
\(\dfrac{1}{10}x-\dfrac{3}{2}=0\)
\(\dfrac{1}{10}x=\dfrac{3}{2}\)
\(x=15\)
\(\dfrac{1}{5}\).x - \(\dfrac{2}{3}\) = \(\dfrac{1}{10}\).x + \(\dfrac{5}{6}\)
⇒ \(\dfrac{1}{5}\).x - \(\dfrac{1}{10}\).x = \(\dfrac{5}{6}\) + \(\dfrac{2}{3}\)
⇒ \(\dfrac{2}{10}\).x - \(\dfrac{1}{10}\).x = \(\dfrac{5}{6}\) + \(\dfrac{4}{6}\)
⇒ \(\dfrac{1}{10}\).x = \(\dfrac{9}{6}\)
⇒ x = \(\dfrac{9}{6}\) : \(\dfrac{1}{10}\)
⇒ x = \(\dfrac{9}{6}\) . 10
⇒ x = \(\dfrac{90}{6}\)
⇒ x = 15
Vậy x = 15