A= \(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+\(\dfrac{1}{10.13}\)+....+\(\dfrac{1}{25.28}\)
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S
10 tháng 5 2023
\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{25.28}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{3}{14}\)
\(=\dfrac{1}{14}\)
#NoSimp
VN
Võ Ngọc Phương
VIP
21 tháng 8 2023
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)
\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=\dfrac{1}{103}\)
\(\Rightarrow x+3=103\)
\(x=103-3\)
\(x=100\)
Vậy x = 100
28 tháng 4 2018
1.
E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)
E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)
E = 1 - \(\dfrac{1}{22}\)
E = \(\dfrac{21}{22}\)
2.
(x - 4)(x - 5) = 0
TH1:
x - 4 = 0 => x = 4
TH2:
x - 5 = 0 => x = 5
Vậy: x = 4 hoặc x = 5
NT
3
2 tháng 8 2018
Ta có : A = 2/ 4.7 + 2/ 7.10 + ... + 2/ 73.76 .
⇒ 3/2 A = 3/2 . ( 2/ 4.7 + 2/ 7.10 + ... + 2/ 73.76 ) .
⇒ 3/2 A = 3/ 4.7 + 3/ 7.10 + ... + 3/ 73.76 .
= 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/ 73 - 1/76 .
= 1/4 - 1/76 .
= 19/76 - 1/76 .
= 9/38 .
Do đó : A = 9/38 : 3/2 .
= 9/38 . 2/3 .
= 3/19 .
Vậy A = 3/19 .
TN
25 tháng 4 2018
\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}< \dfrac{1}{5}\)
=\(\dfrac{3}{3}\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{37.40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)\)
=\(\dfrac{3}{40}< \dfrac{1}{3}\)
27 tháng 4 2017
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
23 tháng 9 2023
`#3107`
`a)`
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=1-\dfrac{1}{2000}\)
\(=\dfrac{1999}{2000}\)
`b)`
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
`c)`
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
VN
Võ Ngọc Phương
VIP
23 tháng 9 2023
b) Sửa đề:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
1 tháng 3 2022
a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)
b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)
c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)
d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
NH
4
VD
5 tháng 11 2017
\(\dfrac{1}{3}\)x(\(\dfrac{3}{1+4}\)+\(\dfrac{3}{4+7}\)+........+\(\dfrac{3}{101+103}\))
\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+.........+\(\dfrac{ }{ }\)\(\dfrac{1}{101}\)-\(\dfrac{1}{103}\))
\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{103}\))
\(\dfrac{1}{3}\)x\(\dfrac{102}{103}\)
\(\dfrac{34}{103}\)
27 tháng 10 2017
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{1}{1}-\dfrac{1}{103}\)
\(=\dfrac{102}{103}\)
\(A=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{6}{28}=\dfrac{2}{28}=\dfrac{1}{14}\)
`3A = 3/(4.7) + 3/(7.10) + .. + 3/(25.28)`
`3A = 1/4 - 1/7 + 1/7 - 1/10 +... + 1/25 - 1/28`
`3A = 3/14`
`A = 1/14.`