\(17,\dfrac{5}{3}.\left(-\dfrac{6}{5}+x\right)-\dfrac{1}{4}.\left(\dfrac{2}{3}-1\right)=-\dfrac{3}{8}\)
\(\dfrac{5}{3}.\left(-\dfrac{6}{5}+x\right)-\dfrac{1}{4}.\left(\dfrac{2}{3}-\dfrac{3}{3}\right)=-\dfrac{3}{8}\)
\(\dfrac{5}{3}.\left(-\dfrac{6}{5}+x\right)-\dfrac{1}{4}.\dfrac{-1}{3}=-\dfrac{3}{8}\)
\(\dfrac{5}{3}.\left(-\dfrac{6}{5}+x\right)-\dfrac{-1}{12}=-\dfrac{3}{8}\)
\(\dfrac{5}{3}.\left(-\dfrac{6}{5}+x\right)=\dfrac{-3}{8}+\dfrac{-1}{12}\)
\(\dfrac{5}{3}.\left(-\dfrac{6}{5}+x\right)=\dfrac{-11}{24}\)
\(\dfrac{-6}{5}+x=-\dfrac{11}{24}:\dfrac{5}{3}\)
\(\dfrac{-6}{5}+x=-\dfrac{11}{24}.\dfrac{3}{5}\)
\(\dfrac{-6}{5}+x=\dfrac{-11}{40}\)
\(x=\dfrac{-11}{40}-\dfrac{-6}{5}\)
\(x=\dfrac{-11}{40}+\dfrac{48}{40}\)
\(x=\dfrac{37}{40}\)
Vậy....
\(18,\dfrac{-3}{4}\left(\dfrac{8}{9}-x\right)+\dfrac{3}{5}=\dfrac{-2}{3}\left(\dfrac{1}{2}\right)\)
\(\dfrac{-3}{4}\left(\dfrac{8}{9}-x\right)+\dfrac{3}{5}=\dfrac{-1}{3}\)
\(\dfrac{-3}{4}\left(\dfrac{8}{9}-x\right)=\dfrac{-1}{3}-\dfrac{3}{5}\)
\(\dfrac{-3}{4}\left(\dfrac{8}{9}-x\right)=\dfrac{-14}{15}\)
\(\dfrac{8}{9}-x=\dfrac{-14}{15}:\dfrac{-3}{4}\)
\(\dfrac{8}{9}-x=\dfrac{-14}{15}.\dfrac{-4}{3}\)
\(\dfrac{8}{9}-x=\dfrac{56}{45}\)
\(x=\dfrac{8}{9}-\dfrac{56}{45}\)
\(x=\dfrac{40}{45}-\dfrac{56}{45}\)
\(x=\dfrac{-16}{45}\)
19: \(\dfrac{2}{3}\left(x-\dfrac{9}{4}\right)=\dfrac{3}{7}\left(7-\dfrac{1}{6}\right)-\dfrac{1}{3}\)
=>\(\dfrac{2}{3}\left(x-\dfrac{9}{4}\right)=3-\dfrac{3}{42}-\dfrac{1}{3}\)
=>\(\dfrac{2}{3}\left(x-\dfrac{9}{4}\right)=\dfrac{8}{3}-\dfrac{3}{42}=\dfrac{109}{42}\)
=>\(x-\dfrac{9}{4}=\dfrac{109}{42}:\dfrac{2}{3}=\dfrac{109}{28}\)
=>\(x=\dfrac{109}{28}+\dfrac{9}{4}=\dfrac{43}{7}\)
20: \(4-\dfrac{2}{3}\left(x-3\right)=2-\dfrac{1}{2}+\dfrac{2}{3}\)
=>\(4-\dfrac{2}{3}x+2=2-\dfrac{1}{2}+\dfrac{2}{3}\)
=>\(6-\dfrac{2}{3}x=\dfrac{12}{6}-\dfrac{3}{6}+\dfrac{4}{6}=\dfrac{13}{6}\)
=>\(\dfrac{2}{3}x=6-\dfrac{13}{6}=\dfrac{36}{6}-\dfrac{13}{6}=\dfrac{23}{6}\)
=>\(x=\dfrac{23}{6}:\dfrac{2}{3}=\dfrac{23}{6}\cdot\dfrac{3}{2}=\dfrac{69}{12}=\dfrac{23}{4}\)
