= 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 :|

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

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

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

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

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

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

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

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

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

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

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

 ~ Hok tốt ~

mình vẫn không hiểu

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)

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

\(=\frac{1}{3}-\frac{1}{15}\)

\(=\frac{4}{15}\)

Chúc bn hok giỏi !!!!!!!!! ^_^

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

\(A=\frac{1}{3}-\frac{1}{19}\)

\(A=\frac{16}{57}\)

Dấu "." là dấu nhân nhá ^^

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{17\cdot19}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{19}\)

\(=\frac{1}{3}-\frac{1}{19}\)

\(=\frac{16}{57}\)

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

\(=\frac{1}{3}-\frac{1}{19}\)

\(=\frac{16}{57}\)

Code : Breacker

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{4}{15}+\frac{9}{5}\)

\(=\frac{31}{15}\)

              Bài làm :

Ta có :

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{31}{15}\)

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

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

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

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

Vậy B = \(\frac{58}{59}\)

Lưu ý: Dấu "." là dấu nhân

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

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

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

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

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

học trường gì

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}\)

\(=\frac{98}{303}\)

Tích mk nha bn !!!! ^_^

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

         \(y=\frac{11}{15}\)

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

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

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

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

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

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

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

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