\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

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

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

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

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

\(D=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{98.100}\)

\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{2}.\dfrac{49}{100}=\dfrac{147}{200}\)

\(D=\dfrac{3}{2\cdot4}+\dfrac{3}{4\cdot6}+\dfrac{3}{6\cdot8}+...+\dfrac{3}{98\cdot100}\\ =\dfrac{3}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\dfrac{49}{100}\\ =\dfrac{147}{200}\)

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

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

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(S=\dfrac{1}{2}-\dfrac{1}{18}-\dfrac{1}{4}+\dfrac{1}{20}\)

\(S=.C.A.S.I.O.\)

A=\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{20\cdot22}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{11}{22}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{5}{11}\)

\(=\dfrac{5}{22}\)

Mk có thể làm wen đc k

Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)

\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow2019n=1009\left(2n+2\right)\)

\(\Leftrightarrow2019n=2018n+2018\)

\(\Leftrightarrow n=2018\)

B = \(\dfrac{12}{\left(2.4\right)^2}+\dfrac{20}{\left(4.6\right)^2}+...+\dfrac{388}{\left(96.98\right)^2}+\dfrac{396}{\left(98.100\right)^2}\)

= \(\dfrac{4^2-2^2}{2^{2^{ }}.4^{2^{ }}}+\dfrac{6^{2^{ }}-4^2}{4^2.6^2}+...+\dfrac{98^2-96^2}{96^2.98^2}+\dfrac{100^2-98^2}{98^2.100^2}\)

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

= \(\dfrac{1}{2^2}-\dfrac{1}{100^2}=\dfrac{1}{4}-\dfrac{1}{100^2}< \dfrac{1}{4}\)

Vậy B < \(\dfrac{1}{4}\)

B =

=

=

=

Vậy B <

Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)

A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)

A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

A=\(\dfrac{98}{99}-\dfrac{49}{100}\)

A=\(\dfrac{4949}{9900}\)

Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)

Vậy A<\(\dfrac{3}{4}\)

Bạn hãy tính \(\dfrac{1}{1.3}+...+\dfrac{1}{98.100}\)= \(\dfrac{4949}{9900}\) sau đo chỉ cần chứng minh nó nhỏ hơn bằng cách quy đồng .

p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)

\(=\dfrac{11}{13}\)