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NQ
1
21 tháng 3 2020
Đặt \(B=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+....+\frac{2}{97\cdot101}\)
\(\Rightarrow2B=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+....+\frac{4}{97\cdot101}\)
\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{97}-\frac{1}{101}\)
\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}\)
\(\Leftrightarrow B=\frac{96}{505}:2\)
DT
0
DT
0
NP
4
D
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
\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)
\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)
\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)
\(S\times4=\frac{1}{5}-\frac{1}{25}\)
\(S\times4=\frac{4}{25}\)
\(S=\frac{1}{25}\)
HT
12 tháng 7 2015
a,\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
b,\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)
=> x = 2008
NP
2
25 tháng 2 2023
\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)
BF
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`
KB
8 tháng 4 2022
Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
9 tháng 1 2022
\(\Leftrightarrow x+2\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)=\dfrac{-74}{45}\)
\(\Leftrightarrow x+2\cdot\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=-\dfrac{74}{45}\)
hay x=-2
VD
23 tháng 4 2022
\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{8}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x=-1\)
\(A=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+...+\frac{2}{55\cdot59}\)
\(A=2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{55}-\frac{1}{59}\right)\)
\(A=2\left(\frac{1}{5}-\frac{1}{59}\right)\)
\(A=2\left(\frac{59}{295}-\frac{5}{295}\right)\)
\(A=2\cdot\frac{54}{295}\)
\(A=\frac{536}{295}\)