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24 tháng 5 2017

\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\)

\(2M=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{995.997}+\dfrac{2}{997.999}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{997}-\dfrac{1}{999}\)

\(=1-\dfrac{1}{999}=\dfrac{998}{999}\)

\(\Rightarrow M=\dfrac{998}{999}.\dfrac{1}{2}=\dfrac{499}{999}\)

\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\\ =\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{995.997}+\dfrac{2}{997.999}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{997}-\dfrac{1}{999}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{999}\right)=\dfrac{1}{2}.\dfrac{998}{999}=\dfrac{499}{999}\)

19 tháng 5 2022

chữ xấu (thông cảm) ;-;

undefined

25 tháng 4 2017

 M = 1 - 1/ 999 = 998/999

25 tháng 4 2017

=> 2M = 2/1*3 + 2/3*5 + ... + 2/995*997  + 2/ 997*999  =  1-1/3 + 1/3 - 1/5 +... + 1/ 995 - 1/997 + 1/997 - 1 / 999 = 1- 1/999 = 998/999 

=> m = 998/999   /  2   =  499/999  (bn tính lại xem nha mk ko có máy tính nên sợ sai)

 Vậy M = ..... 

8 tháng 2 2023

Ta có :

\(\dfrac{1}{1.3}\text{=}2\left(\dfrac{1}{1}-\dfrac{1}{3}\right)\)

\(\dfrac{1}{3.5}\text{=}2\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(\dfrac{1}{5.7}\text{=}2\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)

\(...\)

\(\dfrac{1}{2021.2023}\text{=}2\left(\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(\Rightarrow\) biểu thức chỉ còn :

\(2.1-\dfrac{2}{2023}\text{=}\dfrac{4044}{2023}\)

8 tháng 2 2023

đặt biểu thức trên là A

ta có

2A=2/1.3+2/3.5+...+2/2021.2023

2A=1/1-1/3+1/3-1/5+...+1/2021-1/2023

2A=1/1-1/2023

2A=2022/2023

A=(2022/2023):2

A=1011/2023

19 tháng 3 2023

\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)

\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(B=\dfrac{1}{1}-\dfrac{1}{99}\)

\(B=\dfrac{99}{99}-\dfrac{1}{99}\)

\(B=\dfrac{98}{99}\)

#YVA

22 tháng 3 2023

B=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)

B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)

B=\(\dfrac{98}{99}:2\)

B=\(\dfrac{49}{99}\)

NV
5 tháng 3 2023

\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)

\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)

\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)

\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

\(2H=1-\dfrac{1}{51}\)

\(2H=\dfrac{50}{51}\)

\(H=\dfrac{25}{51}\)

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

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

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

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

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

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

8 tháng 10 2021

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)

7 tháng 7 2023

 

Đặt biểu thức là A

\(2A=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2005-2003}{2003.2005}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}:2=\dfrac{1002}{2005}\)

7 tháng 7 2023

Gọi tổng trên là A. Ta có

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{2005}=\dfrac{2005}{2005}-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

⇒ A= \(\dfrac{2004}{2005}:2=\dfrac{2004}{2005}.\dfrac{1}{2}=\dfrac{1002}{2005}\)

Vậy tổng trên bằng \(\dfrac{1002}{2005}\)

23 tháng 8 2021

a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2003.2004}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}=1-\dfrac{1}{2004}=\dfrac{2003}{2004}\)b)Đặt  \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)\(\Rightarrow A=\dfrac{1002}{2005}\)

a: Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\)

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

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