\(a,\dfrac{2}{3}.\left(x+\dfrac{1}{3}\right)=\dfrac{5}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{5}{2}\)
\(\Rightarrow x=\dfrac{13}{6}\)
Vậy: \(x=\dfrac{13}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}-\dfrac{1}{3}=\dfrac{13}{6}\)
\(\dfrac{2}{3}x+\dfrac{2}{9}=\dfrac{5}{3}\)
\(\dfrac{2}{3}x=\dfrac{5}{3}-\dfrac{2}{9}\)
\(\dfrac{2}{3}x=\dfrac{15}{9}-\dfrac{2}{9}\)
\(x=\dfrac{13}{9}:\dfrac{2}{3}\)
\(x=\dfrac{13}{6}\)
Vậy...
x + \(\dfrac{1}{3}\) = \(\dfrac{5}{3}\) : \(\dfrac{2}{3}\)
x + \(\dfrac{1}{3}\) = \(\dfrac{5}{3}\) x \(\dfrac{3}{2}\)
x + \(\dfrac{1}{3}\) = \(\dfrac{5}{2}\)
x = \(\dfrac{5}{2}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{13}{6}\)
Vậy x = \(\dfrac{13}{6}\)