Tính B = \(\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+....+\frac{9}{1999\cdot2009}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
2
5 tháng 9 2018
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(=\frac{200}{2009}\)
5 tháng 9 2018
Gọi \(B=\frac{9}{19}+A\)
\(A=\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+...+\frac{9}{1999\cdot2009}\)
\(\frac{A}{9}=\frac{1}{19\cdot29}+\frac{1}{29\cdot39}+...+\frac{1}{1999\cdot2009}\)
\(\frac{A\cdot10}{9}=\frac{10}{19+29}+\frac{10}{29\cdot39}+...+\frac{10}{1999\cdot2009}\)
\(\frac{A\cdot10}{9}=\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\)
\(\frac{A\cdot10}{9}=\frac{1}{19}-\frac{1}{2009}\)
\(A=\frac{1791}{38171}\)
\(\Rightarrow B=\frac{1}{19}+\frac{1791}{38171}\)
\(\Rightarrow B=\frac{200}{2009}\)
25 tháng 2 2017
Ta có: \(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
\(\Rightarrow A=\frac{1}{19}+\frac{1791}{38171}\)
\(\Rightarrow A=\frac{200}{2009}\)
Vậy \(A=\frac{200}{2009}.\)
S
25 tháng 7 2018
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)=\frac{9}{10}\cdot\frac{2000}{18081}=\frac{200}{2009}\)
D
1 tháng 9 2018
Ta có: \(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(B=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)\)
\(B=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)
\(B=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(B=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(B=\frac{9}{10}.\frac{2000}{18081}\)
\(B=\frac{200}{2009}\)
Vậy \(B=\frac{200}{2009}\)
17 tháng 2 2017
\(A=\frac{1}{19}+\frac{9}{19.29}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
đến đay bn tự tính nha
S
21 tháng 7 2016
\(\text{Ta có:}\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{2009.2019}\)
\(=\frac{9}{10}.\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+.....+\frac{10}{2009.2019}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2019}\right)\)
\(=\frac{9}{10}.\frac{670}{6057}\)
\(\frac{2000}{2009}\)
\(\frac{200}{2009}\)tính lộn