\(a,\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)

\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=13\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{7.9}\right)\)

\(=13\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(=13.\frac{2}{9}=\frac{26}{9}\)

\(b,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}=\frac{2017}{2018}\)

P/s :Dấu chấm là dấu nhân nha

phần c đâu bn

\(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)

\(=\frac{13.10101}{15.10101}+\frac{13.10101}{35.10101}+\frac{13.10101}{63.10101}+\frac{13.10101}{99.10101}\)

\(=13.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=\frac{13}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{13}{2}.\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=\frac{13}{2}.\frac{8}{33}\)

\(=\frac{52}{33}\)

 \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

Sửa đề chút:

 \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2018}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)

B tự làm nốt nhé

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

a=1/3x5+1/5x7+...+1/2003x2005

a=1x2/3x5x2+1x2/5x7x2+...+1x2/2003x2005x2

a=1/2(2/3x5+2/5x7+...+2/2003x2005)

a=1/2x(1/3-1/5+1/5-1/7+...+1/2003-1/2005)

a=1/2x(1/3-1/2005)

a=1/2x2002/6015

a=1001/6015

A = 1/3.5 + 1/5.7 + 1/7.9 + .... + 1/2003.2005 

2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + .... + 1/2003 - 1/2005

2A = 1/3 - 1/2005 = 2002/6015 

=>A = 1001/6015

\(\frac{1}{2}A=\)\(2\times\left(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{2003\times2005}\right)\)

\(\Leftrightarrow\frac{1}{2}A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{2003\times2005}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{2005}\)

\(\Leftrightarrow\frac{1}{2}A=\frac{2002}{6015}\)

\(\Leftrightarrow A=\frac{1001}{6015}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(\Rightarrow2x=90\)

\(\Rightarrow x=45\)

Đây là tổng của 2 dãy:

\(\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\frac{1}{5\times7\times9}+...+\frac{1}{995\times997\times999}\)(1)

và 

\(\frac{1}{2\times5\times8}+\frac{1}{5\times8\times11}+\frac{1}{8\times11\times14}+...+\frac{1}{1493\times1496\times1499}\)(2)

Dãy số có dạng là tích 3 thừa số, trong đó thừa số thứ 3 hơn thừa số thứ nhất n đơn vị và 2 thừa số cuối của phân số trước là 2 thừa số đầu của phân số sau. Để tính dãy kiểu này cần đưa tử số về hiệu của thừa số thứ 3 và thừa số thứ nhất (hiệu = n):

Vậy nhân dãy thứ nhất với 4:

\(=\frac{4}{1\times3\times5}+\frac{4}{3\times5\times7}+\frac{4}{5\times7\times9}+...+\frac{4}{995\times997\times999}\)

Nhận xét:

Vậy 4 lần tổng dãy 1 là:

\(\frac{1}{1\times3}-\frac{1}{3\times5}+\frac{1}{3\times5}-\frac{1}{5\times7}+...+\frac{1}{995\times997}-\frac{1}{997\times999}\)

\(\frac{1}{1\times3}-\frac{1}{997\times999}\)

Suy ra tổng dãy (1) là \(\left(\frac{1}{3}-\frac{1}{997\times999}\right)\times\frac{1}{4}\)

Làm tương tự tính được tổng dãy (2) là: \(\left(\frac{1}{2\times5}-\frac{1}{1496\times1499}\right)\times\frac{1}{6}\)

Cộng 2 kết quả lại được tổng cần tính

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{48}\)

=> 2x + 3 = 48

=> 2x = 48 - 3

=> 2x = 45

=> x = 45/2

gọi biểu thức đó là A

\(A=\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{2009.2011}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{2008}{6033}\right)\)

\(A=\frac{1004}{6033}\)

mink nghĩ vậy bạn ạ

C.mơn bạn nha ! ^_^

:V Làm sai hết rồi sai ngay từ bước đầu tiên.

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\frac{3}{20}\)

\(=\frac{-11}{12}\)

\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)

= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

= \(-\frac{7}{30}\)

\(=-(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10})\)

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

\(=-\frac{7}{30}\)

~ Hok tốt ~

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)

\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)

\(=\frac{1}{3}-\frac{1}{47.49}\)

\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)