Question

\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

Giúp mình với

Answer 1

Ta có

M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

M = \(\dfrac{1}{3}-\dfrac{1}{99}\)

M = \(\dfrac{32}{99}\)

Vậy M = \(\dfrac{32}{99}\)

Answer 2

\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(M=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(M=\dfrac{1}{3}-\dfrac{1}{99}\)

\(M=\dfrac{32}{99}\)

Answer 3

\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+................+\dfrac{2}{97.99}\)

\(M=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....................+\dfrac{1}{97}-\dfrac{1}{99}\)

\(M=\dfrac{1}{3}-\dfrac{1}{99}\)

\(M=\dfrac{32}{99}\)