Question

a) Tính :

         \(1-\dfrac{1}{2};\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3}-\dfrac{1}{4};\dfrac{1}{4}-\dfrac{1}{5};\dfrac{1}{5}-\dfrac{1}{6}\)

b) Sử dụng kết quả của câu a) để tính nhanh tổng sau :

        \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

Answer 1

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) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

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

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

\(=\dfrac{5}{6}\)

Answer 2