Tính \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}\)
dấu nhân là dấu . nhá mọi người
K
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ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)
\(=\dfrac{8}{12}=\dfrac{2}{3}\)
T
23 tháng 8 2023
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
1 tháng 8 2023
\(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{2021\cdot2022\cdot2023}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}-\dfrac{1}{2022\cdot2023}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4090506}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2045252}{4090506}=\dfrac{1022626}{4090506}=\dfrac{511313}{2045253}\)
KS
2 tháng 8 2023
`1/(1.2.3) + 1/(2.3.4) + ... + 1/(2021 . 2022 .2023)`
`=> 2/(1.2.3) + 2/(2.3.4) + ... + 2/(2021 . 2022. 2023)`
`= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(2021.2022) - 1/(2022 . 2023)`
`= 1/2 - 1/4090506`
`=4090506/8181012 - 2/8181012`
`= 4090504/8181012`
26 tháng 3 2021
Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\dfrac{1}{2}-\dfrac{1}{200}\)
\(=\dfrac{100-1}{200}=\dfrac{99}{200}< \dfrac{100}{200}=\dfrac{1}{2}\)(đpcm)
26 tháng 3 2021
Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\) (Đpcm)
7 tháng 8 2023
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}=\dfrac{10}{11}\)
HQ
Hà Quang Minh
Giáo viên
7 tháng 8 2023
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)
30 tháng 4 2017
A=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2014\cdot2015\cdot2016}=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2014\cdot2015}-\dfrac{1}{2015\cdot2016}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2015}\cdot\dfrac{1}{2016}\right)=\dfrac{1}{4}-\dfrac{1}{2\cdot2015\cdot2016}< \dfrac{1}{4}\)
Vậy A<\(\dfrac{1}{4}\)
---bé hơn hoặc bằng tức là chỉ cần xảy ra 1 khả năng cũng thõa mãn nhé---
L
25 tháng 3 2018
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)
\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)
27 tháng 12 2018
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)
=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)
23 tháng 7 2017
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
23 tháng 7 2017
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\\ =\dfrac{1}{2}-\dfrac{1}{5}\\ =\dfrac{3}{10}\)
\(\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}=\dfrac{1}{2}-\dfrac{1}{3}+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{5}{10}-\dfrac{2}{10}=\dfrac{3}{10}\)