A = \(\frac{1}{5.10}\)+ \(\frac{1}{10.15}\)+ \(\frac{1}{15.20}\)+ ...... + \(\frac{1}{995.1000}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-.....-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
Đặt \(C=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+....+\frac{1}{95.100}\)
\(\Rightarrow C=\frac{1}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+....+\frac{5}{95.100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+....+\frac{1}{95}-\frac{1}{100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)=\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)
\(\Rightarrow1-C=1-\frac{19}{500}=\frac{481}{500}\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có B = 1- 1/5.10 - 1/10.15 -.......- 1/95 .100
=> 5B = 5 -( 5/5.10+5/10.15 +....+ 5/95.100
= > 5B = 5 - ( 1/5 -1/100 )
=> 5B= 481/100
=> B = 481/500
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2}{5.10}+\dfrac{2}{10.15}+...+\dfrac{2}{995.1000}\\ =2\left(\dfrac{1}{5.10}+\dfrac{1}{10.15}+...+\dfrac{1}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{995.1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{995}-\dfrac{1}{1000}\right)\\ =\dfrac{2}{5}\left(\dfrac{1}{5}-\dfrac{1}{1000}\right)\)
\(=\dfrac{2}{5}.\dfrac{199}{1000}\\ =\dfrac{199}{2500}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=3\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{45.50}\right)\)
\(a=\frac{3}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\right)\)
\(a=\frac{3}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\right)\)
\(a=\frac{3}{5}.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(a=\frac{3}{5}\cdot\frac{9}{50}\)
\(a=\frac{27}{250}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(=2.\frac{403}{2020}=\frac{403}{1010}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bai nay de ma hinh nhu day khnog phai cua lop 7 dau ban sai roi
\(A=\frac{1}{5}x\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{995.1000}\right)\)
\(A=\frac{1}{5}x\left(\frac{1}{5}-\frac{1}{1000}\right)\)
\(A=\frac{1}{5}x\frac{199}{1000}\)
\(A=\frac{199}{5000}\)
Nếu muốn thì thử lại :
\(=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+..+\frac{1}{995}-\frac{1}{1000}\right)...\)
\(=\frac{1}{5}\left(1-\frac{1}{1000}\right)=\frac{1}{5}\cdot\frac{995}{1000}\)
tự tính nốt nha