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3 tháng 9 2017

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

\(S=\frac{1}{2}-\frac{1}{10}\)

\(S=\frac{2}{5}\)

3 tháng 9 2017

S = 1 / 2.4 + 1/ 4.6 + 1 / 6.8 + 1 / 8.10

2S = 2 / 2.4 + 2 / 4.6 + 2/ 6.8 + 2 / 8.10

2S = 1 2 - 1 / 4 + 1 / 4 - 1 / 6 + 1 / 6 - 1 / 8 + 1 / 8 - 1 / 10

2S = 1 / 2 - 1 / 10

2S = 2 / 5 

  S = 2 / 5 : 2

  S = 1 / 5 

3 tháng 9 2017

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

\(2S=\frac{1}{2}-\frac{1}{10}\)

\(2S=\frac{2}{5}\)

\(S=\frac{2}{5}:2\)

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

3 tháng 9 2017

S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)

1 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}=\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)

\(S=\frac{29}{45}\)

1 tháng 9 2016

sai roi

1 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\)

\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}-\frac{1}{2}+\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

1 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\right]\)

\(S=\frac{1}{2}.\left[\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\right]\)

\(S=\frac{1}{2}.\left[\left(1-\frac{1}{9}\right)-\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)

\(S=\frac{1}{2}.\left(\frac{8}{9}-\frac{2}{5}\right)\)

\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

8 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

  \(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(=\frac{4}{9}-\frac{1}{5}\)

\(=\frac{11}{45}\)

 

8 tháng 9 2016

Cảm ơn giúp  bài nữa nha !!

7 tháng 9 2016

\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

7 tháng 9 2016

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{4}{9}-\frac{1}{5}\)

\(S=\frac{11}{45}\)

3 tháng 3 2018

\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)

\(S=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}\right)-\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}\right)\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(S=\dfrac{1}{2}-\dfrac{1}{18}-\dfrac{1}{4}+\dfrac{1}{20}\)

\(S=.C.A.S.I.O.\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)

7 tháng 10 2016

Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)

\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)

\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)

\(2A=1-\frac{1}{10}\)

\(2A=\frac{9}{10}\)

\(A=\frac{9}{10}:2=\frac{9}{20}\)

7 tháng 10 2016

=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)

( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)

=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)