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3 tháng 2 2020

Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004.2005}\)

Ta có: \(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004.2005}\)

\(A=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{2004.2005}\right)\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2004.2005}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)

\(A=\frac{2003}{2005}\)

3 tháng 2 2020

bn ơi bn chưa nhân với 2

13 tháng 11 2020

\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{2004.2005}\)

\(\Leftrightarrow2M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{2004.2005}\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{2004.2005}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2004}-\frac{1}{2005}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)

\(=2.\left(\frac{2005}{4010}-\frac{2}{4010}\right)\)

\(=2.\frac{2003}{4010}\)

\(=\frac{2003}{2005}\)

21 tháng 3 2018

\(M=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004\cdot2005}\)

\(M=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004\cdot2005}\)

\(M=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2004\cdot2005}\right)\)

\(M=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2004\cdot2005}\right)\)

\(M=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2004}-\frac{1}{2005}\right)\)

\(M=2\left(\frac{1}{2}-\frac{1}{2005}\right)\)

\(M=2\cdot\frac{2003}{4010}\)

\(M=\frac{2003}{2005}\)

12 tháng 1 2020

\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}\right)=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{9.10}\right)\)

\(=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...-\frac{1}{10}\right)=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{1}{5}-\frac{1}{4}=\frac{-1}{20}\)

12 tháng 1 2020

\(A=\frac{1}{10}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\)

\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\right)\)

\(A=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)

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

\(A=\frac{1}{10}-\left[\left(\frac{1}{4}-\frac{1}{10}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{9}-\frac{1}{9}\right)\right]\)

\(A=\frac{1}{10}-\frac{1}{4}+\frac{1}{10}\)

\(A=\frac{1}{5}-\frac{1}{4}\)

\(A=-\frac{1}{20}\)

13 tháng 8 2015

 ====== 83/88 

8 tháng 3 2017

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2004.2005}\)

\(A=\frac{1}{1.2}=1-\frac{1}{2}\)

\(A=\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(A=1-\frac{1}{2004}\)

\(A=\frac{2003}{2004}\)

Ủng hộ tk Đúng nha mọi người !!! ^^ 

8 tháng 3 2017

\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)\(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\);...; \(\frac{1}{2004.2005}=\frac{1}{2004}-\frac{1}{2005}\)

=> A=\(\frac{1}{1}-\frac{1}{2005}=\frac{2004}{2005}\)