\(\dfrac{1}{99}-\dfrac{1}{97.99}-\dfrac{1}{95.97}-\dfrac{1}{93.95}-...-\dfrac{1}{3.5}-\dfrac{1}{1.3}\)
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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=\(\dfrac{49}{99}\)
HN
2
18 tháng 4 2017
\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{33}{99}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
C
18 tháng 4 2017
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{95}-\dfrac{2}{97}+\dfrac{2}{97}-\dfrac{2}{99}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{99}=\dfrac{64}{99}\)
chúc bạn học tốt
14 tháng 3 2017
Đặt biểu thức cần tính là A, ta có:
A=\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}\right)\)
Làm tg tự với những cái khác là ok
2 tháng 5 2023
Ta có \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2-1+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).
Từ đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\); \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\); \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\); \(1+\dfrac{1}{4.6}=\dfrac{5^2}{4.6}\);...; \(1+\dfrac{1}{2022.2024}=\dfrac{2023^2}{2022.2024}\).
Suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2022.2024}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{2023^2}{2022.2024}\)
\(=\dfrac{2.2023}{2024}\) \(=\dfrac{2023}{1012}\)
OY
30 tháng 7 2021
a) S=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2017.2019}\)
2S=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
2S=\(1-\dfrac{1}{2019}\)
2S=\(\dfrac{2018}{2019}\)
S\(\dfrac{1009}{2019}\)
BD
20 tháng 8 2023
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
=> 2x=98
=> x=49
23 tháng 9 2021
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\Leftrightarrow198x=196x+98\\ \Leftrightarrow2x=98\Leftrightarrow x=49\)
1 tháng 5 2021
A bn lướt xuống dưới mà xem cách làm
nhưng của bn là cho 3 ra ngoài nha
12 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
LT
Lương Thị Vân Anh
CTVHS
12 tháng 3 2023
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
\(\dfrac{1}{99}-\dfrac{1}{97.99}-\dfrac{1}{95.97}-\dfrac{1}{93.95}-...-\dfrac{1}{3.5}-\dfrac{1}{1.3}\\ =\dfrac{1}{99}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}+\dfrac{1}{97.99}\right)\\ \)
\(=\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ \)
\(=\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}+\dfrac{1}{198}=-\dfrac{16}{33}\)
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