Question

[Gia Huy To]

 giúp em bài nay em cảm ơn các quý thầy cô và các bạn

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\left(\dfrac{1}{24}-\dfrac{3}{16}\right):\left(-\dfrac{3}{8}+\dfrac{1}{2}\right)\)

\(=\dfrac{2-9}{48}:\dfrac{-3+4}{8}\)

\(=\dfrac{-7}{48}\cdot\dfrac{8}{1}=\dfrac{-7}{6}\)

b: \(\dfrac{-5}{7}:\left(5-4\dfrac{4}{7}\right)+\left(15\dfrac{1}{3}\cdot\dfrac{-1}{23}\right)\)

\(=\dfrac{-5}{7}:\left(5-\dfrac{32}{7}\right)+\dfrac{46}{3}\cdot\dfrac{-1}{23}\)

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

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

c: \(\left(-1\dfrac{2}{3}\cdot1\dfrac{4}{5}+2\dfrac{5}{6}\right):\left(\dfrac{1}{6}-\dfrac{3}{4}\right)\)

\(=\left(-\dfrac{5}{3}\cdot\dfrac{9}{5}+\dfrac{17}{6}\right):\left(\dfrac{2-9}{12}\right)\)

\(=\left(-3+\dfrac{17}{6}\right):\dfrac{-7}{12}\)

\(=\dfrac{-1}{6}\cdot\dfrac{12}{-7}=\dfrac{12}{42}=\dfrac{2}{7}\)

d: \(\left(\dfrac{1}{3^2}-\dfrac{1}{7^2}\right)\cdot\left(\dfrac{1}{4^2}-\dfrac{1}{7^2}\right)\cdot...\cdot\left(\dfrac{1}{49^2}-\dfrac{1}{7^2}\right)\)

\(=\left(\dfrac{1}{7^2}-\dfrac{1}{7^2}\right)\cdot\left(\dfrac{1}{3^2}-\dfrac{1}{7^2}\right)\cdot...\left(\dfrac{1}{49^2}-\dfrac{1}{7^2}\right)\)

\(=0\left(\dfrac{1}{3^2}-\dfrac{1}{7^2}\right)\cdot...\cdot\left(\dfrac{1}{49^2}-\dfrac{1}{7^2}\right)\)

=0

e: \(\left(1-\dfrac{2}{3}\right)\left(1-\dfrac{2}{5}\right)\cdot...\cdot\left(1-\dfrac{2}{2023}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{3}{5}\cdot\dfrac{5}{7}\cdot...\cdot\dfrac{2021}{2023}=\dfrac{1}{2023}\)