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Toru
7 tháng 11 2023 lúc 11:17

a) \(14\dfrac{2}{5}\cdot\dfrac{7}{8}-6\dfrac{2}{5}\cdot\dfrac{7}{8}\)

\(=\dfrac{7}{8}\cdot\left(14\dfrac{2}{5}-6\dfrac{2}{5}\right)\)

\(=\dfrac{7}{8}\cdot\left(\dfrac{72}{5}-\dfrac{32}{5}\right)\)

\(=\dfrac{7}{8}\cdot\dfrac{40}{5}\)

\(=\dfrac{7}{8}\cdot8=7\)

b) \(\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{-5}{9}-\left(-\dfrac{2021}{2022}\right)^0\)

\(=\dfrac{3}{12}-\dfrac{5}{12}-1\)

\(=-\dfrac{2}{12}-\dfrac{12}{12}\)

\(=-\dfrac{14}{12}=-\dfrac{7}{6}\)

c) \(\dfrac{2}{3}\cdot\left(-6\right)+0,25:1\dfrac{1}{4}\)

\(=-4+\dfrac{1}{4}:\dfrac{5}{4}\)

\(=-4+\dfrac{1}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{-20}{5}+\dfrac{1}{5}=-\dfrac{19}{5}\)

d) \(\left(\left|-0,6\right|+\dfrac{4}{5}\right)\sqrt{\dfrac{9}{49}}+\left(\dfrac{-2}{5}\right)^3\)

\(=\left(0,6+\dfrac{4}{5}\right)\cdot\sqrt{\left(\dfrac{3}{7}\right)^2}+\dfrac{\left(-2\right)^3}{5^3}\)

\(=\dfrac{7}{5}\cdot\dfrac{3}{7}+\dfrac{-8}{125}\)

\(=\dfrac{3}{5}-\dfrac{8}{125}\)

\(=\dfrac{75}{125}-\dfrac{8}{125}=\dfrac{67}{125}\)

\(\text{#}Toru\)

\(a,14\dfrac{2}{5}.\dfrac{7}{8}-6\dfrac{2}{5}.\dfrac{7}{8}=\left(14\dfrac{2}{5}-6\dfrac{2}{5}\right).\dfrac{7}{8}=8.\dfrac{7}{8}=7\\ b,\dfrac{1}{4}+\dfrac{3}{4}.\dfrac{-5}{9}-\left(-\dfrac{2021}{2022}\right)^0\\ =\dfrac{1}{4}+\dfrac{-15}{36}-1\\ =\dfrac{1}{4}-\dfrac{5}{12}-1=\dfrac{3-5-12}{12}=-\dfrac{14}{12}=-\dfrac{7}{6}\)

\(c,\dfrac{2}{3}.\left(-6\right)+0,25:1\dfrac{1}{4}=\dfrac{-12}{3}+\dfrac{1}{4}:\dfrac{5}{4}\\ =-4+\dfrac{1}{5}=-\dfrac{19}{5}\\ d,\left(\left|-0,6\right|+\dfrac{4}{5}\right).\sqrt{\dfrac{9}{49}}+\left(-\dfrac{2}{5}\right)^3\\ =\left(\dfrac{3}{5}+\dfrac{4}{5}\right).\sqrt{\dfrac{3^2}{7^2}}+\dfrac{-8}{125}\\ =\dfrac{7}{5}.\dfrac{3}{7}+\dfrac{-8}{125}=\dfrac{3}{5}+\dfrac{-8}{125}\\ =\dfrac{3.25-8}{125}=\dfrac{67}{125}\)


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