Tính: S=\(\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2017}\)
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HN
3
21 tháng 6 2021
`2/(3.5)+2/(5.7)+....+2/(2015.2017)`
`=1/3-1/5+1/5-1/7+....+1/2016-1/2017`
`=1/3-1/2017=2014/6051`
SI
21 tháng 6 2021
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 5 2022
Đặt tông trên là A
\(\dfrac{2A}{7}=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
\(\Rightarrow A=\dfrac{7.2022}{2.2023}=\dfrac{1011}{289}\)
TC
14 tháng 4 2022
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
5 tháng 5 2022
Đặt tổng trên là A
\(\dfrac{2A}{7}=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}=\)
\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=\)
\(=\dfrac{2022}{2023}\Rightarrow A=\dfrac{7.2022}{2.2023}\)
MH
5 tháng 5 2022
\(A=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{2021}-\dfrac{2}{2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{2023}=\dfrac{4044}{2023}\Rightarrow A=\dfrac{2022}{289}\)
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
20 tháng 9 2021
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{4}{9}-\dfrac{1}{5}\)
\(=\dfrac{11}{45}\)
HH
21 tháng 4 2017
1. Ta có: \(\left|x\right|=7\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy \(x\in\left\{\pm7\right\}\)
2. \(M=\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{13}{39}-\dfrac{3}{39}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{1.10}{2.39}=\dfrac{5}{39}\)
Tick mk vs! Thank nhiều!
21 tháng 4 2017
1. Theo đb ta có: |x|=7
=> Có 2 TH:\(\left\{{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\) \(\in Z\)
Vậy x=7 \(\veebar\) x= -7 ( x\(\in\) Z) thì |x|=7
2. \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Đặt A= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Ta thấy: \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
\(\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{2}{5.7}\)
... \(\dfrac{1}{11}-\dfrac{1}{13}=\dfrac{2}{11.13}\)
=> 2D=2(\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\))
<=> 2D= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
<=>2D=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
<=> 2D= \(\dfrac{1}{3}-\dfrac{1}{13}\)
<=>2D= \(\dfrac{13}{39}-\dfrac{3}{39}\)
<=>2D=\(\dfrac{10}{39}\)
=> D= \(\dfrac{10}{39}:2\)
<=> D= \(\dfrac{10}{39}.\dfrac{1}{2}\)
<=> D=\(\dfrac{5}{39}\)
Vậy D= \(\dfrac{5}{39}\)
_ Chc bn hk tốt_
LS
2
26 tháng 4 2022
\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.100}\)
\(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99.100}+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{3}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=\dfrac{1}{3}-\dfrac{2}{100}+\dfrac{1}{99}\)
\(B=\dfrac{1}{3}-\dfrac{1}{50}+\dfrac{1}{99}\)
Đến đây thì hết tính hợp lý được rồi:v
\(B=\dfrac{34}{99}-\dfrac{1}{50}\)
\(B=\dfrac{1601}{4950}\)
\(S=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2017}\)
\(\dfrac{2}{7}S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(\dfrac{2}{7}S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(\dfrac{2}{7}S=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(\dfrac{2}{7}S=\dfrac{2014}{6051}\)
\(S=\dfrac{4028}{42357}\)
\(S=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2107}\)
\(S=\dfrac{7}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\right)\)
\(S=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(S=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{2017}\right)\)
\(S=\dfrac{7}{2}.\dfrac{2014}{6051}\)
\(S=\dfrac{4028}{42357}\)