a,x.3\(\dfrac{1}{4}\)+(\(\dfrac{-7}{6}\)).x-1\(\dfrac{2}{3}\)=\(\dfrac{5}{12}\)
\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x-\(\dfrac{5}{3}\)=\(\dfrac{5}{12}\)
\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x=\(\dfrac{25}{12}\)
\(\Leftrightarrow\)\(\dfrac{25}{12}\)x=\(\dfrac{25}{12}\)
\(\Leftrightarrow\)x=1
Vậy x=1
b,\(\dfrac{2}{5}\)+\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)
\(\Leftrightarrow\)\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)-\(\dfrac{2}{5}\)
\(\Leftrightarrow\)\(\dfrac{9}{5}\)x-\(\dfrac{111}{50}\)=\(\dfrac{-57}{10}\)
\(\Leftrightarrow\)\(\dfrac{9}{5}\)x=\(\dfrac{-87}{25}\)
\(\Leftrightarrow\)x=\(\dfrac{-29}{15}\)
Vậy x=\(\dfrac{-29}{15}\)
c,\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)+\(\dfrac{5}{9}\)=\(\dfrac{23}{27}\)
\(\Leftrightarrow\)\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)=\(\dfrac{8}{27}\)
\(\Leftrightarrow\)\(\dfrac{7}{18}\)+\(\dfrac{28}{27}\)x=\(\dfrac{8}{27}\)
\(\Leftrightarrow\)\(\dfrac{28}{27}\)x=\(\dfrac{-5}{54}\)
\(\Leftrightarrow\)x=\(\dfrac{-5}{56}\)
Vậy x=\(\dfrac{-5}{56}\)