\(S=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(S=\frac{1}{5}-\frac{1}{95}\)

\(S=\frac{18}{95}\)

Vậy \(S=\frac{18}{95}\)

Giải 

\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(=\frac{1}{5}-\frac{1}{95}\)

\(=\frac{18}{95}\)

Vậy S=\(\frac{18}{95}\)

may bn giai giup minh bai hinh luon voi

\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\frac{8}{33}=\frac{10}{11}.\frac{33}{8}\)

\(=\frac{15}{4}\)

Trả lời:

\(\frac{10}{11}\div\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}\div\frac{8}{33}\)

\(=\frac{10}{11}\times\frac{33}{8}\)

\(=\frac{15}{4}\)

\(\frac{10}{11}:(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11})\)

=\(\frac{10}{11}:(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11})\)

=\(\frac{10}{11}:(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\)

=\(\frac{10}{11}:(\frac{1}{3}-\frac{1}{11})\)

=\(\frac{10}{11}:\frac{8}{33}\)

= \(\frac{10}{11}.\frac{33}{8}\)= \(\frac{15}{4}\)

\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=\frac{10}{11}:\frac{8}{33}\)

\(=\frac{15}{4}\)

Học tốt

\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\frac{8}{33}\)

\(=\frac{10}{11}.\frac{33}{8}\)

\(=\frac{15}{4}\)

\(\frac{10}{11}:\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{10}{11}:\frac{8}{33}=\frac{15}{4}\)

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)

\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)

\(=\frac{99.5}{486}\)

\(=\frac{495}{486}\)

Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)

\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)

\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)

\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)

\(A=66.\frac{13}{27}\)

\(A=\frac{286}{9}\)

sai hay đúng cx ko biết nha

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

\(=1+\frac{2}{3}-\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+....+\frac{2}{15}\)

\(=1+\frac{2}{15}\)

\(=\frac{17}{15}\)

\(A=\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}\)

\(A=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+\frac{2\cdot2}{9\cdot11}\)

\(A=2\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)

\(A=2\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(A=2\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(A=2\cdot\frac{8}{33}\)

\(A=\frac{16}{33}\)

Ta có: 

\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}\)

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

\(A=2\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(A=2\cdot\frac{8}{33}\)

\(A=\frac{16}{33}\)

M =

M = \(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\)

M = \(\frac{2}{1}-\frac{2}{11}\)

M = \(\frac{20}{11}\)

M = 49/515 

tớ ko chép lại đề đâu

 \(\frac{1}{2}M=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{101.103}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{103}\)

\(=\frac{1}{5}-\frac{1}{103}\)

=\(\frac{98}{515}\)

=> \(M=\frac{98}{515}:\frac{1}{2}=\frac{196}{515}\)

Vậy \(M=\frac{196}{515}\)

\(M=\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+.....+\frac{1}{101.103}\)

\(2M=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+....+\frac{2}{101.103}\)

\(2M=\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+.....+\frac{103-101}{101.103}\)

\(2M=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+.....+\frac{1}{101}-\frac{1}{103}\)

\(2M=\frac{1}{5}-\frac{1}{103}\)

\(2M=\frac{88}{515}\)

   \(M=\frac{88}{515}.\frac{1}{2}\)

   \(M=\frac{44}{515}\)

Vậy \(M=\frac{44}{515}\)