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28 tháng 3 2019

 \(A=\frac{-1}{2.4}+\frac{-1}{4.6}+\frac{-1}{6.8}+...+\frac{-1}{98.100}\Leftrightarrow.\)\(-2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\Leftrightarrow.\)

\(-2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\Leftrightarrow.\)

\(-2A=\frac{1}{2}-\frac{1}{100}\Leftrightarrow-2A=\frac{49}{100}\Leftrightarrow A=\frac{-49}{200}.\)

ĐÁP SỐ :   \(A=\frac{-49}{200}.\)

28 tháng 3 2019

\(\frac{-49}{200}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

cho mình nha!

14 tháng 8 2017

Ta có:

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{1}{4,6}+\frac{1}{6.8}+...+\frac{1}{98.100}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)

14 tháng 8 2017

Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)

\(4-2=2;6-4=2;...\)

\(2A=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(2A=\frac{1}{2}-\frac{1}{100}\)

\(2A=\frac{49}{100}\)

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

\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

    \(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)

\(\Rightarrow Q=\frac{49}{40}\)

11 tháng 8 2016

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{49}{100}\)

\(=\frac{49}{200}\)

11 tháng 8 2016

Giúp mk mấy bài nhaEdowa Conan

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

\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)

\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)

\(2.A=\frac{1}{2}-\frac{1}{30}\)

\(2.A=\frac{7}{15}\)

\(A=\frac{7}{15}:2=\frac{7}{30}\)

3 tháng 9 2017

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

\(2S=\frac{1}{2}-\frac{1}{10}\)

\(2S=\frac{2}{5}\)

\(S=\frac{2}{5}:2\)

\(S=\frac{1}{5}\)

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 = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)

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

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

  \(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(=\frac{4}{9}-\frac{1}{5}\)

\(=\frac{11}{45}\)

 

8 tháng 9 2016

Cảm ơn giúp  bài nữa nha !!

20 tháng 6 2020

Đặt BT trên là A

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)

\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)

\(2A=\frac{1}{2}-\frac{1}{102}=\frac{50}{102}\Rightarrow A=\frac{25}{102}\)

Đặt A là biểu thức trên ta có : 

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(=\frac{1}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{50}{102}=\frac{25}{102}\)