Violympic toán 6

NN

a,\(\dfrac{-5}{9}\)+\(\dfrac{5}{9}\):(1\(\dfrac{2}{3}\)-2\(\dfrac{1}{12}\))

b,\(\dfrac{-7}{25}\).\(\dfrac{11}{13}\)+\(\dfrac{-7}{25}\).\(\dfrac{2}{13}\)-\(\dfrac{18}{25}\)

c,(\(\dfrac{5}{7}\).0,6-5:3\(\dfrac{1}{2}\)).(40%-1,4).\(\left(-2\right)^3\)

DB
25 tháng 4 2018 lúc 11:10

a)\(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)

\(=\dfrac{-5}{9}+\dfrac{5}{9}:\left(\dfrac{20}{12}-\dfrac{25}{12}\right)\)

\(=\dfrac{-5}{9}+\dfrac{5}{9}\cdot\dfrac{-12}{5}\)

\(=\dfrac{-5}{9}+\dfrac{-12}{9}\)

\(=\dfrac{-17}{9}\)

b) \(\dfrac{-7}{25}\cdot\dfrac{11}{13}+\dfrac{-7}{25}\cdot\dfrac{2}{13}-\dfrac{18}{25}\)

\(=\dfrac{-7}{25}\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)

\(=\dfrac{-7}{25}\cdot1-\dfrac{18}{25}\)

\(=\dfrac{-25}{25}\)

\(=-1\)

c) \(\left(\dfrac{5}{7}\cdot0,6-5:3\dfrac{1}{2}\right)\cdot\left(40\%-1,4\right)\cdot\left(-2\right)^3\)

\(=\left(\dfrac{3}{7}-\dfrac{10}{7}\right)\cdot\left(0,4-1,4\right)\cdot\left(-8\right)\)

\(=\left(-1\right)\cdot\left(-1\right)\cdot\left(-8\right)\)

\(=-8\)

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