Question

tính nhanh:N= \(\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}+....+\frac{1}{2005x2010}\)=?

 

Answer 1

\(N=\frac{1}{1.5}+\frac{1}{5.10}+...+\frac{1}{2005.2010}=\frac{1}{5}\left(\frac{5}{1.5}+\frac{5}{5.10}+...+\frac{5}{2005.2010}\right)=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)

\(N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{2010}\)

Answer 2

Bài bạn Mạnh Chưa đúng. bạn  Hạnh kiểm tra lại nhé

Answer 3

\(N=\frac{1}{1.5}+\frac{1}{5.10}+....+\frac{1}{2005.2010}\)

\(N=5.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+....+\frac{1}{2005}-\frac{1}{2010}\right)\)

\(N=5.\left(\frac{1}{1}-\frac{1}{2010}\right)\)

\(N=5.\frac{2009}{2010}\)

\(N=\frac{2009}{402}\)

Answer 4

=2009/402

ai thấy đúng thì k nhé

Answer 5

đúng rồi bạn Hạnh chưa đúng

Answer 6

\(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2005.2010}\)

\(4N=\frac{4}{1.5}+\frac{4}{5.10}+\frac{4}{10.15}+...+\frac{4}{2005.2010}\)

\(4N=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\)

\(4N=1-\frac{1}{2010}\)

\(4N=\frac{2010-1}{2010}\)

\(N=\frac{2010-1}{8040}=\frac{2009}{8040}\)