Question

\(\dfrac{1}{10}\)+\(\dfrac{1}{40}\)+\(\dfrac{1}{88}\)+\(\dfrac{1}{154}\)+\(\dfrac{1}{238}\)+\(\dfrac{1}{340}\)

Answer 1

Giải:

Đặt \(A=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)

\(A=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(A=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)

\(A=\dfrac{1}{3}.\dfrac{9}{20}\)

\(A=\dfrac{3}{20}\)

Answer 2

\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

\(=\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+....+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\div3\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\div3\)

\(=\dfrac{9}{20}\div3\)

= \(\dfrac{3}{20}\)

Answer 3

Đặt A=\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

A= \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\) 3A= \(3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\) 3A= \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\) 3A = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\) 3A= \(\dfrac{1}{2}-\dfrac{1}{20}\)

3A= \(\dfrac{9}{20}\)

A = \(\dfrac{9}{20}:3\)

A = \(\dfrac{3}{20}\)

Vậy A= \(\dfrac{3}{20}\)

Chúc bạn học tốt