\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)

\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)

\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)

\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)

\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1}{2}.\frac{1009}{2020}\)

\(\Leftrightarrow A=\frac{1009}{4040}\)

Vậy : \(A=\frac{1009}{4040}\)

Đặt A =\(\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{2018.2010}\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2018.2020}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{2018}-\frac{1}{2020}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2020}\)

\(\Rightarrow A=\frac{1010}{2020}-\frac{1}{2020}\)

\(\Rightarrow A=\frac{1009}{2020}\)

Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=2\cdot\dfrac{505}{1011}\)

\(=\dfrac{1010}{1011}\)

\(A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(A=\frac{1}{2}.\frac{1009}{2020}\)

\(A=\frac{1009}{4040}\)

A=1/2.4+1/4.6+1/6.8+...+1/2018.2020

  =1/2(1/2-1/4+1/4-1/6+...+1/2018-1/2020)

    =1/2(1/2-1/2020)

   =1/2.1009/2020

   =1009/4040

#)Giải :

\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2018.2020}\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\)

     \(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)

     \(2A=\frac{1}{2}-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{6}-\frac{1}{6}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{2018}-\frac{1}{2018}\right)-\frac{1}{2020}\)

     \(2A=\frac{1}{2}-0-0-0-...-0-\frac{1}{2020}\)

     \(2A=\frac{1}{2}-\frac{1}{2020}\)

     \(2A=\frac{1009}{2020}\)

\(\Rightarrow A=\frac{1009}{4040}\)

#)Chúc bn học tốt :D

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

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

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

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

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

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

Gọi tổng cần tính là \(A\)

Ta có: \(A=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{38.40}\)

\(\Rightarrow2A=\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{38.40}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{38}-\dfrac{1}{40}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{40}=\dfrac{19}{40}\)

\(\Rightarrow A=\dfrac{\dfrac{19}{40}}{2}=\dfrac{19}{80}\)

A=1/2.4+1/4.6+........+1/100.102

A=1/2-1/4+1/4-1/6+.......+1/100-1/102

A=1/2-1/102

A=51/102-1/102

A=50/102

A=25/51

a,A =\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{199.200}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{199}-\frac{1}{200}\)

= 1-\(\frac{1}{200}\)

=\(\frac{199}{200}\)

b, B=\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{2018.2020}\)

=3.(\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+..+\frac{1}{2018.2020}\))

=3(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2018}-\frac{1}{2020}\))

= 3.(\(\frac{1}{2}-\frac{1}{2020}\))

=\(\frac{6057}{2020}\)

bạn nhớ tick và theo dõi mk nhé