\(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x=\dfrac{2}{6}-\dfrac{1}{3}=\dfrac{1}{3}-\dfrac{1}{3}=0\)
x+\(\dfrac{1}{3}\)=\(\dfrac{2}{6}\)
=> x=\(\dfrac{2}{6}\)-\(\dfrac{1}{3}\)
=>x=\(\dfrac{2}{6}\)-\(\dfrac{2}{6}\)
x = 0
x+\(\dfrac{1}{3}\) =\(\dfrac{2}{6}\)
x= \(\dfrac{2}{6}\) - \(\dfrac{1}{3}\)
x=0
Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0