B=\(\dfrac{-1}{3}\)+ \(\dfrac{-1}{3.5}\)+ \(\dfrac{-1}{5.7}\)+ \(\dfrac{-1}{7.9}\) +... + \(\dfrac{-1}{99.101}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
K
3
31 tháng 8 2017
Đặt :
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+............+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{98}{303}\)
\(\Leftrightarrow A=\dfrac{49}{303}\)
TV
6
18 tháng 6 2017
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+............+\dfrac{1}{99.101}\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{98}{303}=\dfrac{49}{303}\)
19 tháng 6 2017
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.................+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}.\dfrac{98}{303}\)
\(=\dfrac{49}{303}\)
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: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
8 tháng 10 2021
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)
ND
3 tháng 4 2018
a)
\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}\)
\(=\dfrac{4}{25}\)
b)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
PH
3 tháng 4 2018
a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)
tương tự
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
13 tháng 2 2022
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)
13 tháng 2 2022
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)
TC
14 tháng 4 2022
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)
20 tháng 9 2021
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{4}{9}-\dfrac{1}{5}\)
\(=\dfrac{11}{45}\)
\(B=-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=-\dfrac{50}{101}\)