C=\(\frac{5}{2.4}\)+ \(\frac{5}{4.6}\)+ \(\frac{5}{6.8}\)+......+ \(\frac{5}{2020.2022}\)
K
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Những câu hỏi liên quan
C
2
W
4 tháng 8 2018
\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)
= \(\frac{5}{2}-\frac{5}{100}\)
= \(\frac{49}{50}\)
4 tháng 8 2018
\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
\(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)
\(\Rightarrow Q=\frac{49}{40}\)
NH
1
KT
16 tháng 8 2018
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
KD
16 tháng 8 2018
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
DT
2
8 tháng 5 2016
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}\)
\(=\frac{6}{5}\)
LT
11 tháng 3 2020
\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)
\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)
\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)
VM
3
T
3 tháng 4 2019
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\frac{37}{150}\)
\(=\frac{37}{60}\)
KN
3 tháng 4 2019
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\frac{37}{150}\)
= \(\frac{37}{60}\)
22 tháng 1 2017
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
NA
7
6 tháng 6 2015
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)
6 tháng 6 2015
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)
=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
=\(2\left(1-\frac{1}{1005}\right)\)
=\(2.\frac{1004}{1005}\)
=\(\frac{2008}{1005}\)
P/s: Không biết đúng không nữa, làm đại ^.^
C= 5/2 . (1/2-1/4+1/4-1/6+1/6-1/8+......+1/2020-1/2022)
C= 5/2 . (1/2-1/20220)
C= 5/2. 505/1011
c=2525/2022
\(C=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{2020.2022}\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=\frac{5}{2}.\frac{1010}{2022}=\frac{5050}{4044}=\frac{2525}{2022}\)