\(\dfrac{1}{4.6}+\dfrac{1}{2.6.8}+\dfrac{1}{3.8.10}+\dfrac{1}{4.10.12}\)
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20 tháng 9 2021
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{4}{9}-\dfrac{1}{5}\)
\(=\dfrac{11}{45}\)
13 tháng 2 2022
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
2 tháng 11 2023
Gọi biểu thức trên là A
Ta có:
2A = (\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)).2
2A = \(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{x\left(x+2\right)}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{x+2}\)
mà A = \(\dfrac{1}{10}\)(đề bài)
nên 2A = \(\dfrac{2}{10}\) hay \(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{2}{10}\)
suy ra \(\dfrac{1}{x+2}\) = \(\dfrac{1}{2}\)-\(\dfrac{2}{10}\)=\(\dfrac{3}{10}\)
29 tháng 3 2023
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
3 tháng 3 2018
\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
\(S=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}\right)-\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}\right)\)
\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(S=\dfrac{1}{2}-\dfrac{1}{18}-\dfrac{1}{4}+\dfrac{1}{20}\)
\(S=.C.A.S.I.O.\)
26 tháng 2 2018
Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)
\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow2019n=1009\left(2n+2\right)\)
\(\Leftrightarrow2019n=2018n+2018\)
\(\Leftrightarrow n=2018\)
Đặt \(A=\dfrac{1}{4.6}+\dfrac{1}{2.6.8}+\dfrac{1}{3.8.10}+\dfrac{1}{4.10.12}\)
\(\dfrac{1}{2}A=\dfrac{1}{2.4.6}+\dfrac{1}{4.6.8}+\dfrac{1}{6.8.10}+\dfrac{1}{8.10.12}\)
\(\left(\dfrac{1}{2}A\right).4=\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\)
\(\dfrac{1}{2}.4.A=\dfrac{6-2}{2.4.6}+\dfrac{8-4}{4.6.8}+\dfrac{10-6}{6.8.10}+\dfrac{12-8}{8.10.12}\)
\(2A=\dfrac{6}{2.4.6}-\dfrac{2}{2.4.6}+\dfrac{8}{4.6.8}-\dfrac{4}{4.6.8}+\dfrac{10}{6.8.10}-\dfrac{6}{6.8.10}+\dfrac{12}{8.10.12}-\dfrac{8}{8.10.12}\)
\(2A=\dfrac{1}{2.4}-\dfrac{1}{4.6}+\dfrac{1}{4.6}-\dfrac{1}{6.8}+\dfrac{1}{6.8}-\dfrac{1}{8.10}+\dfrac{1}{8.10}-\dfrac{1}{10.12}\)
\(2A=\dfrac{1}{2.4}-\dfrac{1}{10.12}\)
\(2A=\dfrac{1}{8}-\dfrac{1}{120}\)
\(2A=\dfrac{7}{60}\)
\(A=\dfrac{7}{120}\)