5/2.5 + 5/5.8 + 5/8.11 + ... + 5/ 98.101 = ?
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HH
2
F
5 tháng 10 2020
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{98.101}\)
\(=\frac{5}{2}-\frac{5}{5}+\frac{5}{5}-\frac{5}{8}+....+\frac{5}{98}-\frac{5}{101}\)
\(=\frac{5}{2}-\frac{5}{101}=\frac{495}{202}\)
5 tháng 10 2020
\(\frac{5}{2\times5}+\frac{5}{5\times8}+\frac{5}{8\times11}+...+\frac{5}{98\times101}\)
\(=\frac{5}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{98\times101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{5}{3}\times\frac{99}{202}=\frac{165}{202}\)
15 tháng 8 2016
\(S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)
\(S=\frac{1}{2}-\frac{1}{101}\)
\(S=\frac{99}{202}\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 8 2021
\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{47.50}\)
\(=\frac{5}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{50-47}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{47}-\frac{1}{50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{4}{5}\)
PH
2
NH
26 tháng 2 2023
a) đặt
\(S=1+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\\ 2S=2+\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\\ 2S=2+\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ 2S=2+1-\dfrac{1}{101}\\ 2S=\dfrac{302}{101}\\ S=\dfrac{151}{101}\)
b)
đặt
\(S=\dfrac{1}{2}+\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{98}-\dfrac{1}{101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{101}\\ 3S=\dfrac{201}{101}\\ S=\dfrac{67}{101}\)
TH
Thầy Hùng Olm
Manager
VIP
26 tháng 2 2023
\(2A-1=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(2A-1=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(2A=\dfrac{201}{101}\Rightarrow A=\dfrac{201}{202}\)
DC
3
7 tháng 8 2016
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\frac{99}{202}\)
\(=\frac{66}{101}\)
LH
7 tháng 8 2016
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\)
\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\)
\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)
24 tháng 5 2020
\(M= \dfrac{3^2}{2.5} +\dfrac{3^2}{5.8} +\dfrac{3^2}{8.11}+...+\dfrac{3^2}{98.101}\)
\(M= \) \( \dfrac{9}{2.5} +\dfrac{9}{5.8} +\dfrac{9}{8.11}+...+\dfrac{9}{98.101}\)
\(M=3(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+ \dfrac{3}{98.101})\)
\(M= 3(\dfrac{1}{2} -\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11})\)
\(M= 3(\dfrac{1}{2}-\dfrac{1}{11})\)
\(M=3(\dfrac{11}{22}- \dfrac{2}{22})\)
\(M=3.\dfrac{9}{22}\)
\(M=\dfrac{27}{22}\)
SK
2
T
8 tháng 6 2019
#)Giải :
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{99.101}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow3A=\frac{99}{202}\)
\(\Leftrightarrow A=\frac{33}{202}\)
8 tháng 6 2019
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)
NT
1
6 tháng 9 2020
a) 2/2.5 + 2/5.8 + 2/8.11 + ... + 2/x(x+3) = 7/23
3/2.5 + 3/5.8 + 3/8.11 + ... + 3/x(x+3) = 21/46
1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x+1 = 21/46
1/2 - 1/x+1 = 21/46
=> 1/x+1 = 1/23
=> x + 1 = 23
=> x = 22
Vậy x = 22.
b) 3/4 . x - 1/5 = 7/4 . x + 11/5
3/4 . x - 7/4 . x = 1/5 + 11/5
x (3/4 - 7/4) = 12/5
-x = 12/5
x = -12/5
Vậy x = -12/5.
TC
2 tháng 5 2022
\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{98}=\dfrac{49}{98}-\dfrac{1}{98}=\dfrac{48}{98}=\dfrac{24}{49}\)
NA
2 tháng 5 2022
\(A=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)=\dfrac{1}{3}\cdot\dfrac{24}{49}=\dfrac{8}{49}\)
\(\frac{5}{2.5}+\frac{5}{5.8}+......+\frac{5}{98.101}\)
\(=\frac{5}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+........+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)=\frac{5}{3}.\frac{99}{202}\)
\(=\frac{5.33}{202}=\frac{165}{202}\)