^{\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{57.59}\)}

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{59}\)

\(B=1-\frac{1}{59}\)

\(B=\frac{59}{59}-\frac{1}{59}=\frac{58}{59}\)

Vậy B = \(\frac{58}{59}\)

Lưu ý: Dấu "." là dấu nhân

\(B=\frac{2}{1.3}+\frac{1}{3.5}+\frac{2}{5.7}+...+\frac{1}{57.59}\)

\(B=1.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{57}-\frac{5}{59}\right)\)

\(B=1.\left(1-\frac{1}{59}\right)\)

\(B=1.\frac{58}{59}\)

\(B=\frac{58}{59}\)

= 1-1/3 + 1/3-1/5+.......+1/97-1/99

=  1 - 1/99

= 98/99

sao lại là 1- 1/3 + 1/3 -1/5 + ...... 1/97 - 1/99 hả bạn :|

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{95.97}\)\(+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

\(y=\frac{2}{3}.\frac{11}{10}\)

\(y=\frac{22}{30}\)

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

         \(\frac{10}{11}.y=\frac{2}{3}\)

                    \(y=\frac{11}{15}\)

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{11-9}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

         \(y=\frac{2}{3}:\frac{10}{11}\)

         \(y=\frac{11}{15}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{13.15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

mik đã trả lời rồi mà , sao chưa hiện ra ????

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}+\frac{2}{13\times15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}=\frac{14}{15}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(S=\frac{1}{2}.\frac{2016}{2017}\)

\(S=\frac{1008}{2017}< \frac{1}{2}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(2S=1-\frac{1}{2017}< 1\)

=> 2S < 1 

=> S < \(\frac{1}{2}\)(đpcm)

S = \(\frac{1}{2}\). \(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2017}\right)\)= \(\frac{1}{2}.\frac{2016}{2017}\)= \(\frac{1008}{2017}< \frac{1008}{2016}=\frac{1}{2}\)
(=) \(S< \frac{1}{2}\)(đpcm)

           p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)

<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019

<=> p = 1/3 - 1/2019

<=> p = 224/673

\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)

\(=\frac{112}{673}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{2015.2017}+\frac{2}{2017.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2019}\right)=\frac{1}{2}.\frac{224}{673}=\frac{112}{673}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)

\(=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

 ~ Hok tốt ~

mình vẫn không hiểu

\(\frac{2^2}{1x3}\)x \(\frac{4^2}{3x5}\)x \(\frac{6^2}{5x7}\) x \(\frac{8^2}{7x9}\)

= \(\frac{4}{3}\)x \(\frac{16}{15}\)x \(\frac{36}{35}\)x \(\frac{64}{63}\)

= \(1.486077098\)