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27 tháng 4 2021

Đặt tổng trên là A

\(4A=\frac{5-1}{1.5}+\frac{9-5}{5.9}+\frac{13-9}{9.13}+...+\frac{25-21}{21.25}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)

\(4A=1-\frac{1}{25}\Rightarrow A=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

27 tháng 4 2021

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}=\frac{1}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{21.25}\right)\)

\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\right)\)

\(=\frac{1}{4}\left(1-\frac{1}{25}\right)=\frac{1}{4}.\frac{24}{25}=\frac{6}{25}< \frac{1}{4}\)

10 tháng 5 2017

\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)

\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(=2100\left(1-\frac{1}{25}\right)\)

\(=2100\cdot\frac{24}{25}\)

\(=2016\)

10 tháng 5 2017

\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\frac{24}{25}\)

\(A=2016\)

29 tháng 3 2017

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{17\cdot21}< 1\)

\(A=\dfrac{4}{4}\cdot\left(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{17\cdot21}\right)< 1\)

\(A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}< 1\)

\(A=1-\dfrac{1}{21}< 1\) (đúng) (đpcm).

29 tháng 3 2017

Đề sai

11 tháng 4 2017

\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)

=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)

=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)

=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)

11 tháng 4 2017

\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)

\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)

\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)

\(S\times4=\frac{1}{5}-\frac{1}{25}\)

\(S\times4=\frac{4}{25}\)

\(S=\frac{1}{25}\)

2 tháng 8 2016

4/1.5+4/5.9+4/9.13+....+4/21.25

=1-1/5+1/5-1/9+1/9-1/13+......+1/21-1/25

=1-1/25

=24/25

Tích đúng cho mình nha

2 tháng 8 2016

\(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{21}-\frac{1}{25}\)

\(=1-\frac{1}{25}=\frac{24}{25}\)

25 tháng 2 2023

\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)

25 tháng 2 2023

`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`

`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`

`= 1/5 - 1/25`

`= 4/25`

 

24 tháng 8 2017

\(S1=\dfrac{5}{10.11}+\dfrac{5}{11.12}+.............+\dfrac{5}{14.15}\)

\(\Leftrightarrow S1=5\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+...............+\dfrac{1}{14.15}\right)\)

\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+.............+\dfrac{1}{14}-\dfrac{1}{15}\right)\)

\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{15}\right)\)

\(\Leftrightarrow S1=5.\dfrac{1}{30}=\dfrac{1}{6}\)

\(S2=\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}+........+\dfrac{1}{21.25}\)

\(\Leftrightarrow4S_2=\dfrac{4}{5.9}+\dfrac{4}{9.13}+..............+\dfrac{4}{21.25}\)

\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+............+\dfrac{1}{21}-\dfrac{1}{25}\)

\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{25}\)

\(\Leftrightarrow4S_2=\dfrac{4}{25}\)

\(\Leftrightarrow S_2=\dfrac{16}{25}\)

26 tháng 4 2018

- A ở trên giữa các phân số là dấu " + " nha mấy bạn !

7 tháng 3 2017

\(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+......+\frac{3}{21.25}\)

\(=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{21.25}\right)\)

\(=\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{21}-\frac{1}{25}\right)\)

\(=\frac{3}{4}\left(1-\frac{1}{25}\right)\)

\(=\frac{3}{4}.\frac{24}{25}\)

\(=\frac{18}{25}\)

7 tháng 3 2017

\(4A=3-\frac{1}{5}+\frac{3}{5}-\frac{3}{9}+\frac{3}{9}-\frac{3}{13}+...+\frac{3}{21}-\frac{3}{25}\)\(\frac{3}{25}\)

\(4A=3-\frac{3}{25}\)

\(4A=\frac{72}{25}\)

\(A=\frac{18}{25}\)

k minh ha

24 tháng 2 2022

bạn sửa số cuối tử là 4 nhé 

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{401}-\dfrac{1}{405}=1-\dfrac{1}{405}=\dfrac{404}{405}\)

24 tháng 2 2022

\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{401.405}\\ =1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{401}-\dfrac{1}{405}\\ =1-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{401}-\dfrac{1}{401}\right)-\dfrac{1}{405}\\ =1-0-0-....-0-\dfrac{1}{405}\\ =1-\dfrac{1}{405}\\ =\dfrac{404}{405}\)