\(\frac{2^2}{2.4}+\frac{2^2}{4.6}+...+\frac{2^2}{26.28}\)

= \(2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{26.28}\right)\)

= \(2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{28}\right)\)

= \(2.\left(\frac{1}{2}-\frac{1}{28}\right)\)

= \(2.\frac{13}{28}\)

= \(\frac{13}{14}\)

\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{26.28}\right)=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{28}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{28}\right)\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{13.14}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{13}-\frac{1}{14}\)

\(=1-\frac{1}{14}=\frac{13}{14}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\)

\(A=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\right)\)

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

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

\(A=\frac{1}{2}.\frac{7}{15}\)

\(A=\frac{7}{30}\)

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

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

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

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

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

\(A=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)

\(2A=\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+\frac{2}{4\times6}+\frac{2}{5\times7}+\frac{2}{6\times8}+\frac{2}{7\times9}+\frac{2}{8\times10}\)

\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\)

\(2A=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\)

\(2A=\frac{58}{45}\)

\(A=\frac{58}{45}\div2\)

\(A=\frac{29}{45}\)

\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{10}\)

\(=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}=\frac{58}{45}\)

\(A=\frac{29}{45}\)

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

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

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

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

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

Giúp mk mấy bài nhaEdowa Conan

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

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

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

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

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

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

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

cho mình nha!

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 : 

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

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

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

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

Kudo Shinichi mik sẽ biết ơn ai giải cho mik nhìu nhìu

Đề kiểu j v

Ta có

=\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{8.10}\right)\)

=\(\frac{4}{3}.\frac{9}{8}....\frac{81}{80}\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{9.9}{8.10}\)

=\(\frac{2.3....9}{1.2....8}.\frac{2.3....9}{3.4....10}\)

=\(9.\frac{2}{10}\)

=\(\frac{9}{5}\)

A = \(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2012.2014}\)

A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2012}-\frac{1}{2014}\right)\)

A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014}\right)\)

A = \(\frac{1}{2}.\frac{503}{1007}\)

A = \(\frac{503}{2014}\)

2A=\(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2014-2012}{2012.2014}\)

\(2A=\frac{4}{2.4}-\frac{2}{2.4}+\frac{6}{4.6}-\frac{4}{4.6}+...+\frac{2014}{2012.2014}-\frac{2012}{2012.2014}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)

\(2A=\frac{1}{2}-\frac{1}{2014}=\frac{503}{1007}\Rightarrow A=\frac{503}{2014}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2012.2014}\)

\(\Rightarrow2A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2012.2014}\right)\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2012.2014}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2014}\)

\(\Rightarrow2A=\frac{503}{1007}\)

\(\Rightarrow A=\frac{503}{2014}\)