K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

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

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

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

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

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

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

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

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

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 !!

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

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

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

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

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

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

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

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

1 tháng 8 2018

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

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

\(2A=\dfrac{1}{2}\left[1-\dfrac{1}{9}+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\right]\)

\(2A=\dfrac{1}{2}\left(\dfrac{8}{9}+\dfrac{2}{5}\right)\)

\(2A=\dfrac{1}{2}.\dfrac{58}{45}\)

\(2A=\dfrac{29}{45}\)

\(A=\dfrac{29}{45}:2=\dfrac{29}{90}\)

1 tháng 8 2018

A= 1/1.3+1/2.4+1/3.5+...+1/7.9+1/8.10

A = (1/1.3+1/3.5 + 1/5.7 + 1/7.9) + (1/2.4 + 1/4.6 + 1/6.8 + 1/8.10)

A = 1/2. (2/1.3+2/3.5 + 2/5.7 + 2/7.9) + 1/2. (2/2.4 + 2/4.6 + 2/6.8 + 2/8.10)

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

A = 1/2.8/9 + 1/2.2/5

A = 4/9 + 1/5

A = 20/45 + 9 /45

A = 29/45

9 tháng 5 2017

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

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

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

2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{7.9}\)

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

2A = \(1-\frac{1}{9}=\frac{8}{9}\)

A = \(\frac{8}{9}:2=\frac{4}{9}\)

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

2B = \(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{8.10}\)

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

2B = \(\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

B = \(\frac{2}{5}:2=\frac{1}{5}\)

Thay A và B vào S ta được:

\(S=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)

9 tháng 5 2017

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

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

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

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

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

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

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

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