\(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2x-3}{6}=\dfrac{1}{5}\)
\(\Leftrightarrow5\cdot\left(2x-3\right)=6\)
\(\Leftrightarrow10x=6+15=21\)
\(\Leftrightarrow x=\dfrac{21}{10}\)
Ta có: \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{10}\)
\(\Leftrightarrow x=\dfrac{3.7}{10}=\dfrac{21}{10}=2,1\)
Vậy x=2,1
x/3 - 1/2 = 1/5
x/10 - 5/10 = 2/10
x/10. = 2/10+5/10
x/10. = 7/10. => x = 7
\(\Rightarrow\) \(\dfrac{x}{3}\) = \(\dfrac{1}{5}\) + \(\dfrac{1}{2}\)
\(\Rightarrow\) \(\dfrac{x}{3}\) = \(\dfrac{2}{10}\) + \(\dfrac{5}{10}\)
\(\Rightarrow\) \(\dfrac{x}{3}\) = \(\dfrac{7}{10}\)
Ta có: \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}=\dfrac{7}{10}\)
hay \(x=\dfrac{21}{10}\)
Vậy: \(x=\dfrac{21}{10}\)
\(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{3.7}{10}=\dfrac{21}{10}\)