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12 tháng 5 2017

Ta có: \(2S=\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{7\cdot9}+\dfrac{2}{8\cdot10}\)

\(\Leftrightarrow2S=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{10}\)

\(\Leftrightarrow2S=1+\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{58}{45}\)

\(\Rightarrow S=\dfrac{29}{45}\)

13 tháng 5 2017

Ta có:

\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}+\dfrac{1}{8.10}\)

\(=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\right)\) \(+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\right)\)

Đặt \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{7.9}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{7.9}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)=\dfrac{4}{9}\)

Đặt \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{8.10}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{8.10}\right)\)

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

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

\(\Rightarrow S=A+B=\dfrac{4}{9}+\dfrac{1}{5}=\dfrac{29}{45}\)

Vậy \(S=\dfrac{29}{45}\)

26 tháng 4 2016

S=(1/1.3+1/3.5+.....+1/7.9) + (1/2.4+1/4.6+....+1/8.10)

2S=1/2.(1-1/9+(1/2-1/10))

2S=1/2.(8/9 + 2/5)

2S=1/2.58/45

2S=29/45

S=29/45:2

S=29/90

26 tháng 4 2016

S=(1/1.3+1/3.5+.....+1/7.9) + (1/2.4+1/4.6+....+1/8.10)

2S=1/2.(1-1/9+(1/2-1/10))

2S=1/2.(8/9 + 2/5)

2S=1/2.58/45

2S=29/45

S=29/45:2

S=29/90

10 tháng 8 2015

S= 1/1.3+ 1/2.4+1/3.5+....+!/7.9+1/8.10
=1/2(1-1/3 +1/2-1/4 +1/3-1/5 +...+ 1/7-1/9 + 1/8-1/10)
=1/2(1+1/2-1/9-1/10)
=....

10 tháng 8 2015

\(S=\frac{1}{1.3}+.....+\frac{1}{8.10}\)

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

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\)

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

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

\(S=\frac{9}{10}:2\)

\(S=\frac{9}{20}\)

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 !!

23 tháng 4 2016

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

S = 1/2 ( 1 -1/3 +1/2-1/4+......+ 1/8-1/10)

S = 1/2(1+1/2-1/9-1/10)

S= 29/45

8 tháng 6 2016

Bạn nói cô giáo sửa đề thành: 

Tính tổng S=1/1.3+1/2.4+1/3.5+.....+1/\(7\).9+1/8.10 

chứ không tổng S lẻ lắm, chẳng ai muốn tính cả.

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

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

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

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

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

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

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