\(\frac{2}{2x4}\)+\(\frac{2}{4x6}\)+\(\frac{2}{6x8}\)+\(\frac{2}{8x10}\)+\(\frac{2}{10x12}\)+\(\frac{2}{12x14}\)
K
Khách
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Những câu hỏi liên quan
10 tháng 5 2020
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
10 tháng 5 2020
\(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}+\frac{2}{10\times12}+\frac{2}{12\times14}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}+\frac{1}{10}-\frac{1}{10}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}=\frac{3}{7}\)
3 tháng 3 2017
Đặt : \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{10}-\frac{1}{12}\)
\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow A=\frac{5}{12}\)
HI
2
SM
26 tháng 9 2018
\(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}\)
\(=\frac{2}{2}-\frac{2}{4}+\frac{2}{4}-\frac{2}{6}+\frac{2}{6}-\frac{2}{8}+\frac{2}{8}-\frac{2}{10}\)
\(=\frac{2}{2}-\frac{2}{10}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
NH
3
6 tháng 5 2017
Ta gọi biểu thức đó là A
Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức ta có
\(\frac{4}{2.4}=2.\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(\frac{4}{4.6}=2.\left(\frac{1}{4}-\frac{1}{6}\right)\)
\(....................\)
\(\frac{4}{18.20}=2.\left(\frac{1}{18}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{18}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{9}{20}\right)=\frac{18}{20}\)
Ai thấy đúng thì ủng hộ nha !!!
ND
5
HT
12 tháng 7 2015
\(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).y=\frac{1}{3}\)
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}:\frac{1}{2}=\frac{2}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{2}{3}\)
\(\frac{2}{5}.y=\frac{2}{3}\)
=> \(y=\frac{2}{3}:\frac{2}{5}\)
=>\(y=\frac{3}{5}\)
IM
31 tháng 8 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
4 tháng 3 2017
=1/2-1/4+1/4-1/6+1/6-1/8+...+1/2014-1/2016
=1/2-1/2016
=1007/2016
TK va ket ban nhe
D
5 tháng 3 2016
\(\frac{1}{2x4}+\frac{1}{4x6}+...+\frac{1}{96x98}+\frac{1}{98x199}=\frac{2}{2x4}+\frac{2}{4x6}+...+\frac{2}{99x100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
12 tháng 3 2016
A x2 = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{100}\)
A x2 = \(\frac{49}{100}\)
A = \(\frac{49}{200}\)
SC
3
20 tháng 10 2017
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Vậy: \(A=\frac{49}{100}\)
20 tháng 10 2017
Ta có:\(2A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{98.100}\right)\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)
Vậy giá trị của A là \(\frac{49}{200}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)