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a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)
b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)
\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)
\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)
\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)
\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)
\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)
\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)
\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)
\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)
\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)
\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A=1-\frac{1}{2010}\)
\(A=\frac{2009}{2010}\)
Ta có:
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{1}{4,6}+\frac{1}{6.8}+...+\frac{1}{98.100}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(4-2=2;6-4=2;...\)
\(2A=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(2A=\frac{1}{2}-\frac{1}{100}\)
\(2A=\frac{49}{100}\)
mk chỉ tiềm đc bài i hệt bài của bn
https://olm.vn/hoi-dap/detail/99402078680.html
\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(K=2\times\frac{502}{1005}\)
\(K=\frac{1004}{1005}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{10}{33}:3\)
\(F=\frac{10}{99}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}\)
\(I=\frac{2009}{2010}\)
a, \(A=\frac{6}{10.11}+\frac{6}{11.12}+\frac{6}{12.13}+...+\frac{6}{69.70}\)
\(A=\frac{6}{10}-\frac{6}{11}+\frac{6}{11}-\frac{6}{12}+\frac{6}{12}-\frac{6}{13}+...+\frac{6}{69}-\frac{6}{70}\)
\(A=\frac{6}{10}-\frac{6}{70}\)
\(A=\frac{18}{35}\)
b, \(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
\(B=\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(B=2.\frac{1009}{2020}\)
\(B=\frac{1009}{1010}\)
Chúc bạn học tốt 
Hơi thắc mắc câu B cậu oi!!!Gỉai thích cho mk vs ạ!!Thanks
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
A=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
\(\frac{1}{2}\)A= \(\frac{1}{2}.\left(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\right)\)
\(\frac{1}{2}A\)= \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\)
\(\frac{1}{2}A\)= \(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2020-2018}{2018.2020}\)
\(\frac{1}{2}A\)= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(\frac{1}{2}A\)= \(\frac{1}{2}-\frac{1}{2020}\)
\(\frac{1}{2}A=\frac{1009}{2020}\)
\(A=\frac{1009}{2020}:\frac{1}{2}\)
\(A=\frac{1009}{1010}\)
a) Ta có
A= 4/2*4+4/4*6+....+4/2018*2020
=> A= 2*(2/2*4+2/4*6+...+2*(2018*2020)
=> A= 2*(1/2-1/4+1/4-1/6+...+1/2018-1/2020)
=> A= 2*(1/2-1/2020)
=> A= 2* 1009/2020
=> A= 1009/1010
b) B= 1/18+1/54+1/108+...+1/990
=> B= 3/3*(1/18+1/54+1/108+..+1/990)
=> B= 1/3*( 3/3*6+3/6*9+...+3/30*33)
=> B= 1/3*(1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33)
=> B= 1/3*( 1/3-1/33)
=> B=1/3*10/33
=> B=10/99