Bài 3:
a) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}=\dfrac{4}{3}-\dfrac{-4}{3}=\dfrac{8}{3}\)
b) \(2\dfrac{1}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+0,4.5\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+\dfrac{2}{5}.5\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\left(\dfrac{2}{3}+2\right)\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\left[\dfrac{-3}{2}+\dfrac{8}{3}\right]\)
\(=\dfrac{7}{3}-\dfrac{1}{3}.\dfrac{7}{6}\)
\(=\dfrac{7}{3}-\dfrac{7}{18}\)
\(=\dfrac{35}{18}\)
c) \(\left(20+9\dfrac{1}{4}\right):2\dfrac{1}{4}=\left(20+\dfrac{37}{4}\right):\dfrac{9}{4}=\dfrac{117}{4}:\dfrac{9}{4}=13\)
d) \(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)
\(=\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\)
\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{32}{5}\)
\(=10-\dfrac{32}{5}\)
\(=\dfrac{18}{5}\)
e) \(\dfrac{32}{15}:\left(-1\dfrac{1}{5}+1\dfrac{1}{3}\right)=\dfrac{32}{15}:\left(\dfrac{-6}{5}+\dfrac{4}{3}\right)=\dfrac{32}{15}:\dfrac{2}{15}=16\)
g) \(0,2.\dfrac{15}{36}-\left(\dfrac{2}{5}+\dfrac{2}{3}\right):1\dfrac{1}{5}\)
\(=\dfrac{1}{5}.\dfrac{5}{12}-\dfrac{16}{15}:\dfrac{6}{5}\)
\(=\dfrac{1}{12}-\dfrac{8}{9}\)
\(=\dfrac{-29}{36}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{8}{15}+0,25\right).\dfrac{24}{27}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{8}{9}\)
\(=\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{8}{9}\)
\(=\dfrac{7}{5}-\dfrac{94}{135}\)
\(=\dfrac{19}{27}\)
g) \(5:\left(4\dfrac{3}{4}-1\dfrac{25}{28}\right)-1\dfrac{3}{8}:\left(\dfrac{3}{8}+\dfrac{9}{20}\right)\)
\(=5:\left(\dfrac{19}{4}-\dfrac{53}{28}\right)-\dfrac{11}{8}:\dfrac{33}{40}\)
\(=5:\dfrac{20}{7}-\dfrac{5}{3}\)
\(=\dfrac{7}{4}-\dfrac{5}{3}\)
\(=\dfrac{1}{12}\)
\(a\)) \(\dfrac{3}{4}.\dfrac{16}{9}-\dfrac{7}{5}:\dfrac{-21}{20}\)
\(=\dfrac{4}{3}-\dfrac{7}{5}.\dfrac{-20}{21}\)
\(=\dfrac{4}{3}-\dfrac{-4}{3}\)
\(=\dfrac{8}{3}\)
\(=2\dfrac{2}{3}\)