TB

a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)

b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\)

c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)

d) 40% \(-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\)

PT
14 tháng 5 2021 lúc 14:17

\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)

\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)

\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)

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MY
14 tháng 5 2021 lúc 14:22

a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)

\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)

\(=\dfrac{2}{5}+-2\)

\(=\dfrac{2}{5}+\dfrac{-10}{5}\)

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

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a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{5}+-2=\dfrac{-8}{5}\)  

b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)=\dfrac{19}{5}-\dfrac{9}{4}-\dfrac{9}{5}=\left(\dfrac{19}{5}-\dfrac{9}{5}\right)-\dfrac{9}{4}=2-\dfrac{9}{4}=\dfrac{-1}{4}\) 

c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\) 

\(=\dfrac{-3}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\) 

\(=\dfrac{-3}{5}.1+\dfrac{3}{5}\) 

\(=\dfrac{-3}{5}+\dfrac{3}{5}\) 

\(=0\) 

d) \(40\%-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\) 

\(=\dfrac{2}{5}-\dfrac{12}{7}:\dfrac{3}{7}+\dfrac{9}{5}\) 

\(=\dfrac{2}{5}-4+\dfrac{9}{5}\) 

\(=\dfrac{-9}{5}\)

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MY
14 tháng 5 2021 lúc 14:29

b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\)

\(=\dfrac{19}{5}-\left(\dfrac{9}{4}+\dfrac{9}{5}\right)\)

\(=\dfrac{19}{5}-\dfrac{9}{4}-\dfrac{9}{5}\)

\(=\dfrac{19}{5}-\dfrac{9}{5}-\dfrac{9}{4}\)

\(=2-\dfrac{9}{4}\)

\(=\dfrac{8}{4}-\dfrac{9}{4}\)

\(=\dfrac{-1}{4}\)

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MY
14 tháng 5 2021 lúc 14:35

c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)

\(=\dfrac{-3}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\)

\(=\dfrac{-3}{5}.1+\dfrac{3}{5}\)

\(=\dfrac{-3}{5}+\dfrac{3}{5}\)

\(=0\)

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