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26 tháng 4 2018

1/3+1/15+1/35+1/63+1/99+1/143

=1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13

=1/2x(2/1x3+2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)

=1/2x(1/1-1/3+1/3-1/5+...+1/11-1/13)

=1/2x[1/1-(1/3-1/3)-(1/5-1/5)-...-(1/11-1/11)-1/13]

=1/2x(1/1-0-0-...-0-1/13)

=1/2x(1/1-1/13)

=1/2x12/13

=6/13

19 tháng 9 2018

không lẽ nào bằng nhau được đâu vì

\(\left(\frac{1}{30}+\frac{1}{42}\right)+\left(\frac{1}{56}+\frac{1}{72}\right)+\left(\frac{1}{90}+\frac{1}{110}\right)+\left(\frac{1}{132}+\frac{1}{156}\right)+\left(\frac{1}{182}+\frac{1}{210}\right)=\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\)

Hình như bn làm sai ấy. Chứ bài này sách của mk giải đc mà.

30 tháng 8 2015

\(\frac{5}{3}+\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+\frac{5}{143}\)

\(=\frac{5}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{11\cdot13}\right)\)

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

\(=\frac{5}{2}\cdot\left(1-\frac{1}{13}\right)\)

\(=\frac{5}{2}\cdot\frac{12}{13}\)

\(=\frac{30}{13}\)

30 tháng 8 2015

\(\frac{5}{3}+\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+\frac{5}{143}\)

\(=5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)

\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

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

\(=\frac{5}{2}\left(1-\frac{1}{13}\right)\)

\(=\frac{5}{2}.\frac{12}{13}\)

\(=\frac{30}{13}\)

2 tháng 7 2020

\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)

\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)+\left(1-\frac{1}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)

\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

2 tháng 7 2020

Mk sửa lại 1 tí nha dòng thứ 5 :

\(A=6-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}\left(1-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}.\frac{12}{13}\)

\(=6-\frac{6}{13}=\frac{72}{13}\)

Mong bn bỏ qua nha

25 tháng 7 2019

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

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

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

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

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

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

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

25 tháng 7 2019

câu c hình như sai đề hả bn

1 tháng 7 2015

\(\frac{2^2}{15}+\frac{2^2}{35}+\frac{2^2}{63}+\frac{2^2}{99}+\frac{2^2}{143}=2\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\right)=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)=2\cdot\left(\frac{1}{3}-\frac{1}{13}\right)=2\cdot\frac{10}{39}=\frac{20}{39}\)

1 tháng 7 2015

\(=2\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

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

              \(=2\left(1-\frac{1}{13}\right)=2.\frac{12}{13}=\frac{24}{13}\)

23 tháng 4 2018

2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

      = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

      =\(1-\frac{1}{15}=\frac{14}{15}\)

\(\Rightarrow S=\frac{7}{15}\)

23 tháng 4 2018

a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)

A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)

A=2(1-1/15)

A=1/2.14/15

A=7/15

  • kết quả bg 120/17
                      
15 tháng 7 2015

Gọi dãy là A ta có :

A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13

A = 1/2 . ( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 )

A = 1/2 . ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 )

A = 1/2 . ( 1/3 - 1/13 )

A = 1/2 . 10/39

A = 5/39

15 tháng 7 2015

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

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

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

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

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

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