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26 tháng 3 2020

M =1/1.2+1/2.3+1/3.4+.......+1/2019.2020

=1-1/2+1/2-1/3+1/3-1/4+.......-1/2019+1/2019-1/2020

=1-1/2020

=2019/2020

3 tháng 5 2022

\(\text{#}HaimeeOkk\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)

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

\(A=1-\dfrac{1}{2020}\)

\(A=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

24 tháng 8 2019

\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\) 

\(\Rightarrow a=\frac{2020}{2019}\)

24 tháng 8 2019

=.> 1-1/2+1/2-1/3+.......+1/2019-1/2020=1/x

=>1-1/2020=1/x

=>2019/2020=1/x

=>2019x=2020

=>x=2020/2019

    k nha

 giúp mk lên 300sp

14 tháng 6 2020

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2019.2020}\)

\(\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(\frac{1}{4}A=1-\frac{1}{2020}=\frac{2019}{2020}\)

\(\Rightarrow A=\frac{2019}{2020}:\frac{1}{4}=\frac{2019}{505}\)

Vậy \(A=\frac{2019}{505}.\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2B=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)

\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)

Vậy \(B=\frac{4949}{19800}.\)

14 tháng 6 2020

\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2019\cdot2020}\)

\(A=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\right)\)

\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(A=4\left(1-\frac{1}{2019}\right)=4\cdot\frac{2018}{2019}\)

Đến đây tự tính

\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

Số hơi bị dữ nên tính nốt nhé

21 tháng 1 2016

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(S=1-\frac{1}{10}=\frac{9}{10}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\\ =\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}\)

1 tháng 8 2021

Sai :(

ko phải \(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)

Mà là \(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)

5 tháng 5 2019

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{12.13}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{12}-\frac{1}{13}\)

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

\(=\frac{12}{13}\)

5 tháng 5 2019

1/1.2+1/2.3+...+1/12.13

=1-1/2+1/2-1/3+...+1/12-1/13

=1-1/13

=12/13

1 tháng 8 2017

1/1.2+1/2.3+1/3.4+......+1/2003.2004=1/1-1/2+1/2-1/3+1/3-1/4+......+1/2003-1/2004

                                                          =1/1-1/2004

                                                          =2003/2004

1 tháng 8 2017

1/1.2+1/2.3+1/3.4+.......1/2003.2004

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

6 tháng 1 2017

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)\(=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

6 tháng 1 2017

thank you