\(a.x+\dfrac{3}{7}=\dfrac{2}{5}+\dfrac{3}{10}\)
\(x+\dfrac{3}{7}=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}-\dfrac{3}{7}\)
\(x=\dfrac{19}{70}\)
\(b.\dfrac{19}{20}-x=\dfrac{8}{5}-\dfrac{3}{4}\)
\(\dfrac{19}{20}-x=\dfrac{17}{20}\)
\(x=\dfrac{19}{20}-\dfrac{17}{20}\)
\(x=\dfrac{2}{20}=\dfrac{1}{10}\)
\(c.\dfrac{7}{5}-\dfrac{x}{4}=\dfrac{13}{20}\)
\(x=\dfrac{7}{5}-\dfrac{13}{20}\)
\(x=\dfrac{15}{20}=\dfrac{3}{4}\)
=> Vậy x = 3