Tính nhanh:
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\)
K
Khách
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NT
6
KK
4
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
V
2
NK
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
H1
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
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2023
\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)
\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)
\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)
\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2H=1-\dfrac{1}{51}\)
\(2H=\dfrac{50}{51}\)
\(H=\dfrac{25}{51}\)
20 tháng 9 2021
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{4}{9}-\dfrac{1}{5}\)
\(=\dfrac{11}{45}\)
8 tháng 10 2021
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
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 \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)\(\Rightarrow A=\dfrac{1002}{2005}\)
23 tháng 8 2021
a: Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(=\dfrac{2003}{2004}\)
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\)
\(2M=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{995.997}+\dfrac{2}{997.999}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{997}-\dfrac{1}{999}\)
\(=1-\dfrac{1}{999}=\dfrac{998}{999}\)
\(\Rightarrow M=\dfrac{998}{999}.\dfrac{1}{2}=\dfrac{499}{999}\)
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\\ =\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{995.997}+\dfrac{2}{997.999}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{997}-\dfrac{1}{999}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{999}\right)=\dfrac{1}{2}.\dfrac{998}{999}=\dfrac{499}{999}\)