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1 tháng 1 2016

\(\frac{2000}{2009}\)

1 tháng 1 2016

\(\frac{200}{2009}\)tính lộn

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 \(A=\frac{200}{2009}.\)

25 tháng 7 2018

\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)

\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)=\frac{9}{10}\cdot\frac{2000}{18081}=\frac{200}{2009}\)

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

17 tháng 2 2017

cảm ơn nhg mình lm đc rùi

21 tháng 7 2016

\(\text{Ta có:}\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{2009.2019}\)

\(=\frac{9}{10}.\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+.....+\frac{10}{2009.2019}\right)\)

\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2019}\right)\)

\(=\frac{9}{10}.\frac{670}{6057}\)