\(\dfrac{10}{3}\cdot\dfrac{1}{2}-\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\dfrac{1}{5}=\dfrac{3}{5}:\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{10}{3}\cdot\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{3}{5}\cdot2\\ \Leftrightarrow\dfrac{1}{2}x=1\\ \Leftrightarrow x=2\)
Vậy x=2.
\(\dfrac{10}{3}.\dfrac{1}{2}-\left(\dfrac{1}{2}.x-\dfrac{1}{3}\right)+\dfrac{1}{5}=\dfrac{3}{5}:\dfrac{1}{2}\\ \dfrac{5}{3}-\left(\dfrac{1}{2}.x-\dfrac{1}{3}\right)+\dfrac{1}{5}=\dfrac{6}{5}\\ \dfrac{5}{3}-\left(\dfrac{1}{2}.x-\dfrac{1}{3}\right)=\dfrac{6}{5}-\dfrac{1}{5}=1\\ \dfrac{1}{2}.x-\dfrac{1}{3}=\dfrac{5}{3}-1=\dfrac{2}{3}\\ \dfrac{1}{2}.x=\dfrac{2}{3}+\dfrac{1}{3}=1\\ x=1:\dfrac{1}{2}=2\\ Vậy:x=2\)
\(\dfrac{10}{3}.\dfrac{1}{2}-\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\dfrac{1}{5}=\dfrac{3}{5}:\dfrac{1}{2}\)
\(\Rightarrow\dfrac{5}{3}-\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\dfrac{1}{5}=\dfrac{3}{5}.\dfrac{2}{1}\)
\(\Rightarrow\dfrac{5}{3}-\dfrac{1}{2}x+\dfrac{1}{3}+\dfrac{1}{5}=\dfrac{6}{5}\)
\(\Rightarrow\dfrac{5}{3}+\dfrac{1}{3}-\dfrac{1}{2}x=\dfrac{6}{5}-\dfrac{1}{5}\)
\(\Rightarrow2-\dfrac{1}{2}x=1\Rightarrow\dfrac{1}{2}x=1\Rightarrow x=1:\dfrac{1}{2}=1.\dfrac{2}{1}=2\)
