\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(\frac{2}{5}.7+\frac{2}{7}.9+\frac{2}{9}.11+...+\frac{2}{49}.51\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{46}{255}\)

\(=\frac{23}{255}\)

\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)

\(\Rightarrow2 \left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)

\(\Rightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{49}-\frac{1}{51}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)

Vì biểu thức đã được nhân 2 nên giá trị của biểu thức là:

\(\frac{46}{255}:2=\frac{23}{255}\)

\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\frac{46}{255}\)

\(=\frac{23}{255}\)

\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(M=\frac{1}{3}-\frac{1}{13}\)

\(M=\frac{10}{39}\)

\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(M=\frac{1}{2}.\frac{10}{39}\)

\(M=\frac{5}{39}\)

tk mk nha bn

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

=\(\frac{1}{2}x\left(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{2015x2017}\right)\)

=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{2017}\right)\)

=\(\frac{1}{2}x\frac{2012}{10085}\)

=\(\frac{1006}{10085}\)

=1/5-1/2017

tk mình nha

thanks

ta có : công thức 1/n.(n+k) = 1/k.(1/n-1/k)

=> 1/5.7+1/7.9.........+1/2015.2017

= 1/2.(1/5-1/7)+1/2.(1/7-1/9)...............+1/2.(1/2015-1/2017)

= 1/2.(1/5-1/7+1/7-1/9+.........+1/2015-1/2017)

=1/2.(1/5-1/2017)

=1/2.2012/10085

=1006/10085 

( có thể là kết quả sai nhưng cách làm là đug rùi đấy ) * ^_^ *

\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)

<=> \(2.\frac{1}{2}-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=\frac{8}{5}-2x\)

<=> \(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)=\frac{8}{5}-2x\)

<=> \(1-\left(1-\frac{1}{11}\right)-\frac{8}{5}=-2x\)

<=> \(-\frac{83}{55}=-2x\)

<=> \(x=\frac{83}{110}\)

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a

Tính :

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\Rightarrow\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{5}{15}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)

bằng.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................mấy mình cũng ko biết