=5/2(1/4-1/6+1/6-1/8+...+1/208-1/300)

=5/2(1/4-1/300)

=5/2.37/150=37/60

\(\frac{3}{6.8}+\frac{3}{8.10}+.......+\frac{3}{198.200}\)

\(=\frac{3}{2}.\left(\frac{2}{6.8}+\frac{2}{8.10}+........+\frac{2}{198.200}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+........+\frac{1}{198}-\frac{1}{200}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{200}\right)\)

\(=\frac{3}{2}.\frac{97}{600}=\frac{97}{400}\)

\(3.\left(\frac{2}{6.8}+\frac{2}{8.10}+....+\frac{2}{198.200}\right).\frac{1}{2}\)

=\(3.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{198}{200}\right).\frac{1}{2}\)

=\(3.\left(\frac{1}{6}-\frac{1}{200}\right).\frac{1}{2}\)

=.\(3.\frac{97}{600}.\frac{1}{2}\)=97/400

\(\frac{3}{6.8}+\frac{3}{8.10}+......+\frac{3}{198.200}\)

\(=\frac{3}{2}.\left(\frac{2}{6.8}+\frac{2}{8.10}+........+\frac{2}{198.200}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.......+\frac{1}{198}-\frac{1}{200}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{200}\right)\)

\(=\frac{3}{2}.\frac{97}{600}=\frac{97}{400}\)

\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\) 

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\) 

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\) 

\(=\frac{5}{2}\cdot\frac{37}{150}\) 

\(=\frac{37}{60}\)

\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)

= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)

= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)

= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)

= \(\frac{5}{2}.\frac{37}{150}\)

= \(\frac{37}{60}\)

Ta đặt biểu thức trên là A

\(\frac{1}{2}\)A=\(\frac{2}{4.6}\)+\(\frac{2}{6.8}\)+...........+\(\frac{2}{298.300}\)

\(\frac{1}{2}\)A=1/4-1/6+1/6-1/8+..............+1/298-1/300

\(\frac{1}{2}\)A=1/4-1/300

1/2A=74/300

A=74/300:1/2

A=37/75

\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)

\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)

\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)

\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)

\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+...+\frac{5}{298\cdot300}\)

\(=\frac{5}{2}\cdot\left(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\right)\)

\(=\frac{5}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)

\(=\frac{5}{2}\cdot\left(\frac{1}{4}-\frac{1}{300}\right)\)

\(=\frac{37}{60}\)

\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)

\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)

\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)

mình ko chép lại đề nhé, sửa 2014 + 2016 thành 2014.2016

\(A=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)=2\left(\dfrac{2016-2}{6032}\right)=\dfrac{2.2018}{6032}=\dfrac{4036}{6032}=\dfrac{1009}{1508}\)

\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2014.2016}\)

\(=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2014.2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)\)

\(=2.\dfrac{1007}{2016}=\dfrac{1007}{1008}\)

Giải:

Sửa đề 4/2014+2016 ➜ 4/2014.2016

A=4/2.4+4/4.6+4/6.8+...+4/2014.2016

A=2.(2/2.4+2/4.6+2/6.8+...+2/2014.2016)

A=2.(2/2-2/4+2/4-2/6+2/6-2/8+...+2/2014-2/2016)

A=2.(2/2-2/2016)

A=2.1007/1008

A=1007/504

Chúc bạn học tốt!