tính
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2018.2020}\)
K
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LT
11 tháng 3 2020
\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)
\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)
\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)
NN
2
MA
9 tháng 9 2016
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\)
\(A=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}.\frac{7}{15}\)
\(A=\frac{7}{30}\)
9 tháng 9 2016
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)
\(2.A=\frac{1}{2}-\frac{1}{30}\)
\(2.A=\frac{7}{15}\)
\(A=\frac{7}{15}:2=\frac{7}{30}\)
1 tháng 8 2019
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 
6 tháng 10 2019
Hơi thắc mắc câu B cậu oi!!!Gỉai thích cho mk vs ạ!!Thanks
HH
4
3 tháng 9 2017
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
\(2S=\frac{1}{2}-\frac{1}{10}\)
\(2S=\frac{2}{5}\)
\(S=\frac{2}{5}:2\)
\(S=\frac{1}{5}\)
HN
3 tháng 9 2017
S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)
AN
3
19 tháng 3 2017
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{38}-\frac{1}{40}\)
=\(\frac{1}{2}-\frac{1}{40}\)
=\(\frac{19}{40}\)
19 tháng 3 2017
= 2 *[1/2 * 1/4 +1/4 *1/6 +1/6*1/8+...+1/38*1/40
=2*[1/2 - 1/40]
=2 * (-19/40)
= -380
8 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
MA
7 tháng 9 2016
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
HT
7 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
NT
1
6 tháng 3 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
cho mình nha!
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1}{2}.\frac{1009}{2020}\)
\(\Leftrightarrow A=\frac{1009}{4040}\)
Vậy : \(A=\frac{1009}{4040}\)