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\(E=\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{27}{12.39}=\frac{1}{3}-\frac{1}{12}+\frac{1}{12}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)

 

7 tháng 9 2015

lộn bạn đăng từng câu thôi

13 tháng 9 2020

\(-\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-\frac{1}{2001\cdot2000}-...-\frac{1}{2\cdot1}\) 

\(=-1\left(\frac{1}{1\cdot2}+...+\frac{1}{2000\cdot2001}+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\) 

\(=-1\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\) 

\(=-1\left(1-\frac{1}{2003}\right)\) 

\(=-1\left(\frac{2003}{2003}-\frac{1}{2003}\right)\)              

\(=-1\cdot\frac{2002}{2003}\) 

\(=-\frac{2002}{2003}\)

14 tháng 7 2018

\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)

    \(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

     \(=-\left(\frac{1}{2003}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2001}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)

      \(=-\left(\frac{1}{2003}-1\right)=-\left(-\frac{2002}{2003}\right)=\frac{2002}{2003}\)

Vậy ....

21 tháng 6 2016

\(=\frac{3}{3.13}+\frac{3}{13.23}+...+\frac{3}{1993.2003}\)

\(=\frac{1}{10}.\left(1-\frac{3}{13}+\frac{3}{13}-\frac{3}{23}+...+\frac{3}{1993}-\frac{3}{2003}\right)\)

\(=\frac{1}{10}.\left(1-\frac{3}{2003}\right)\)

\(=\frac{1}{10}.\frac{2000}{2003}\)

\(=\frac{200}{2003}\)

21 tháng 6 2016

Đặt \(A=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(\Rightarrow A=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(\Rightarrow A=3\left(\frac{1}{3.13}+\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\)

\(\Rightarrow A=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\right)\)

\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)

\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)\)

\(\Rightarrow A=\frac{3}{10}.\left(\frac{2003}{6009}-\frac{3}{6009}\right)\)

\(\Rightarrow A=\frac{3}{10}.\frac{2000}{6009}\)

\(\Rightarrow A=\frac{200}{2003}\)

29 tháng 7 2021

 \(N=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+..+\frac{10}{1993.2003}\right)\)

\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)

\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)=\frac{3}{10}.\frac{2000}{6009}=\frac{200}{2003}\)

29 tháng 7 2021

\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{3}{13.23}\)\(+\)\(\frac{3}{23.33}\)\(+...+\)\(\frac{3}{1993.2003}\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}.\frac{1990}{26039}\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{597}{26039}\)

\(N=\)\(\frac{200}{2003}\)

30 tháng 8 2019

Bạn tham khảo ở lcik này ! Mình mới trả lời ở đó !

Câu hỏi của Nguyễn Thị Ngọc Ánh - Toán lớp 7 - Học toán với OnlineMath

https://olm.vn/hoi-dap/detail/228829251573.html

24 tháng 6 2017

\(\frac{1}{13}+\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\)

\(=\frac{1}{13}+\left[\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13\cdot23}+\frac{1}{23\cdot33}+...+\frac{1}{1993\cdot2003}\right]\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right]\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{2003}\right]\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\cdot\frac{1990}{26039}\right]\)

\(=\frac{1}{13}+\frac{597}{26039}\)

\(=\frac{200}{2003}\)

24 tháng 6 2017

Đặt A= 1/13 + 3/13.23 + 3/ 23.33 + ... + 3/1993.2003 

A- 1/13 = 3/13.23 + 3/ 23.33 + ... + 3/1993.2003 

10/3 ( A-1/3) =  10/3. (3/13.23 + 3/ 23.33 + ... + 3/1993.2003) 

10/3A - 10/9 = 10/13.23 + 10/ 23.33 + ... + 10/1993.2003 

10/3A - 10/9  = 1/13 - 1/23 + 1/23 - 1/33 +...+ 1/1993- 1/2003

10/3A = 1/13 - 1/2003 + 10/9

10/3 A= ? 

đến đây bn tự làm nha

10/3A - 10/9 = 1/13