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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
23 tháng 5 2022
a: \(=\left(-\dfrac{3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{5}{2}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)
\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{5}{3}-\dfrac{8}{13}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{-413}{156}=\dfrac{-2891}{312}\)
b: \(=-\dfrac{3}{5}:\left(\dfrac{-2-5}{30}\right)+\dfrac{3}{5}:\left(-\dfrac{1}{3}-\dfrac{16}{15}\right)\)
\(=\dfrac{-3}{5}\cdot\dfrac{-30}{7}+\dfrac{3}{5}:\dfrac{-5-16}{15}\)
\(=\dfrac{3}{5}\cdot\dfrac{30}{7}+\dfrac{3}{5}\cdot\dfrac{-5}{7}\)
\(=\dfrac{3}{5}\cdot\dfrac{25}{7}=\dfrac{15}{7}\)
c: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=\left(3-5-6\right)+\left(\dfrac{-1}{4}+\dfrac{7}{4}-\dfrac{3}{2}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\dfrac{6}{5}\)
\(=-8+1+\dfrac{6}{5}=-7+\dfrac{6}{5}=\dfrac{-35+6}{5}=\dfrac{-29}{5}\)
\(-\dfrac{2}{3}+\dfrac{6}{5}:\dfrac{2}{3}+\left(\dfrac{-2}{15}\right)=\dfrac{2}{3}+\left(-\dfrac{6}{5}\right):\dfrac{2}{3}+\left(\dfrac{-2}{15}\right)=\dfrac{2}{3}:\left(-\dfrac{6}{5}+\dfrac{-2}{15}\right)=\dfrac{2}{3}:\left(-\dfrac{18}{15}+\dfrac{-2}{15}\right)=\dfrac{2}{3}:\dfrac{-20}{15}=\dfrac{2}{3}X\dfrac{15}{-20}=\dfrac{1}{1}x\dfrac{5}{-10}=\dfrac{5}{-10}=\dfrac{1}{-2}\)
\(\dfrac{-2}{3}+\dfrac{6}{5}:\dfrac{2}{3}+\dfrac{-2}{15}\)
\(=\dfrac{-2}{3}+\dfrac{6}{5}\times\dfrac{3}{2}-\dfrac{2}{15}\)
\(=\dfrac{-2}{3}+\dfrac{9}{5}-\dfrac{2}{15}\)
\(=\dfrac{-10}{15}+\dfrac{27}{15}-\dfrac{12}{15}\)
\(=\dfrac{5}{15}=\dfrac{1}{3}\)