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a) \(1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5-4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6-5}{30}=\dfrac{1}{30}\)
b) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+-\dfrac{1}{6}\)\(=1+-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
bài 1.
A) \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}=\left(\dfrac{-5}{9}-\dfrac{3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)=\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)
B)\(-\dfrac{5}{13}+\left(\dfrac{3}{5}+\dfrac{3}{1}-\dfrac{4}{10}\right)=-\dfrac{5}{13}+3+\left(\dfrac{3}{5}-\dfrac{4}{10}\right)=-\dfrac{5+3.13}{13}+\dfrac{1}{5}=-\dfrac{207}{65}\)C)\(\dfrac{5}{17}-\dfrac{9}{15}-\dfrac{2}{-17}+\dfrac{-3}{15}=\left(\dfrac{5}{17}+\dfrac{2}{17}\right)+\left(\dfrac{-9}{15}+\dfrac{-3}{15}\right)=\dfrac{7}{17}-\dfrac{12}{15}=-\dfrac{33}{85}\)D)\(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{-1}{9}=\left(\dfrac{1}{9}+\dfrac{-1}{9}\right)+\left(\dfrac{3}{6}+\dfrac{1}{2}\right)-\dfrac{12}{17}=1-\dfrac{12}{17}=\dfrac{5}{17}\)
