a, \(\dfrac{7}{18}x-\dfrac{2}{3}=\dfrac{5}{18}\)
\(\dfrac{7}{18}x=\dfrac{5}{18}+\dfrac{2}{3}\)
\(\dfrac{7}{18}x=\dfrac{17}{18}\)
\(x=\dfrac{17}{18}\div\dfrac{7}{18}\)
\(x=\dfrac{17}{7}\)
a) \(\dfrac{7}{18}\).x-\(\dfrac{2}{3}\)=\(\dfrac{5}{18}\)
\(\dfrac{7}{18}\).x =\(\dfrac{5}{18}\)+\(\dfrac{2}{3}\)
\(\dfrac{7}{18}\).x = \(\dfrac{17}{18}\)
x = \(\dfrac{17}{18}\) :\(\dfrac{7}{18}\)
x =\(\dfrac{17}{7}\)
b)\(\dfrac{4}{9}\) - \(\dfrac{7}{8}\).x =\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{4}{9}\)-\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{10}{9}\)
x =\(\dfrac{10}{9}\) : \(\dfrac{7}{8}\)
x =\(\dfrac{80}{63}\)
c)\(\dfrac{1}{6}\)+\(\dfrac{-5}{7}\): \(\dfrac{-7}{18}\)
\(\dfrac{1}{6}\)+\(\dfrac{90}{49}\)
\(\dfrac{589}{294}\)
a) \(\dfrac{7}{18}\).x -\(\dfrac{2}{3}\)= \(\dfrac{5}{18}\)
\(\dfrac{7}{18}\). x= \(\dfrac{5}{18}\)+\(\dfrac{2}{3}\)
\(\dfrac{7}{18}\). x= \(\dfrac{17}{18}\)
x= \(\dfrac{17}{18}\): \(\dfrac{7}{18}\)
x= \(\dfrac{68}{63}\)
Vậy x= \(\dfrac{68}{63}\)
b) \(\dfrac{4}{9}-\dfrac{7}{8}\). x=\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}-x=\)\(\dfrac{4}{9}-\dfrac{-2}{3}\)
\(\dfrac{7}{8}\)- x= \(\dfrac{10}{9}\)
x=\(\dfrac{7}{8}-\dfrac{10}{9}\)
x=\(\dfrac{-17}{72}\)
Vậy x= \(\dfrac{-17}{72}\)
c) \(\dfrac{1}{6}\)+ \(\dfrac{-5}{7}\): \(\dfrac{-7}{18}\)
=\(\dfrac{1}{6}\)+ \(\dfrac{90}{49}\)
=\(\dfrac{589}{294}\)