Question

[Za Warudo]

Giúp mình với:

a) A=\(\dfrac{1}{3}\)-\(\dfrac{3}{5}\)+\(\dfrac{5}{7}\)-\(\dfrac{7}{9}\)+\(\dfrac{9}{11}\)-\(\dfrac{11}{13}\)+\(\dfrac{13}{15}\)+\(\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)

b) B= \(\dfrac{1}{9.10}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-...-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)

Answer 1

[OH-YEAH^^]

a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

A=0+0+0+...+0+\(\dfrac{13}{15}\)

A=\(\dfrac{13}{15}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)