\(\left(\dfrac{1}{3}+x\right):\dfrac{2}{3}=1\)
\(\left(\dfrac{1}{3}+x\right)=1\times\dfrac{2}{3}\)
\(\dfrac{1}{3}+x=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}-\dfrac{1}{3}\)
\(x=\dfrac{1}{3}\left(tm\right)\)
Vậy ..................
\(\left(\dfrac{1}{3}+x\right)\div\dfrac{2}{3}=1\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{1}{2}=1\)
\(\Leftrightarrow\dfrac{3}{2}x=1-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}\div\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}\cdot\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{2}{6}\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(x=\dfrac{1}{3}\)
\(\left(\dfrac{1}{3}+x\right):\dfrac{2}{3}=1\)
\(\left(\dfrac{1}{3}+x\right)=1.\dfrac{2}{3}\)
\(\left(\dfrac{1}{3}+x\right)=\dfrac{2}{3}\)
\(\dfrac{1}{3}+x=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}-\dfrac{1}{3}\)
\(x=\dfrac{1}{3}\)
(\(\dfrac{1}{3}\)+x):\(\dfrac{2}{3}\)=1
\(\dfrac{1}{3}\)+x =1x\(\dfrac{2}{3}\)
\(\dfrac{1}{3}\)+x =\(\dfrac{2}{3}\)
x =\(\dfrac{2}{3}\)-\(\dfrac{1}{3}\)
x =\(\dfrac{1}{3}\)