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Khách
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16 tháng 5 2016
S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)
= 1/3 . (1/2 - 1/20)
= 1/3 . 9/20
= 3/20
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
16 tháng 5 2016
\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)
\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(S=\frac{9}{20}:3=\frac{3}{20}\)
26 tháng 4 2018
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)
A = \(\frac{1}{2}-\frac{1}{98}\)
A = \(\frac{24}{49}\)
Vậy A = \(\frac{24}{49}\)
~~~
#Sunrise
KT
26 tháng 4 2018
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(=\frac{1}{3}.\frac{24}{49}=\frac{8}{49}\)
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
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)
LN
31 tháng 1 2019
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
28 tháng 3 2017
\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
NN
2
26 tháng 8 2020
\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{10300}=\frac{1}{x}\)
=> \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{100\cdot103}=\frac{1}{x}\)
=> \(\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{100\cdot103}\right)=\frac{1}{x}\)
=> \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{1}{x}\)
=> \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{103}\right)=\frac{1}{x}\)
=> \(\frac{101}{618}=\frac{1}{x}\)
=> \(101x=618\)
=> \(x=\frac{618}{101}\)
Vậy : ...
F
26 tháng 8 2020
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{10300}=\frac{1}{x}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{100.103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{x}.3=\frac{101}{206}\)
\(\Rightarrow\frac{1}{x}=\frac{101}{618}\)
\(\Rightarrow x=\frac{618}{101}\)
NT
2
14 tháng 2 2016
\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)
\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}\left(\frac{34}{68}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)
TM
12 tháng 5 2017
A=...
<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)
NH
0
Ta có: 3S = 3/2.5 + 3/5.8 + ... + 3/47.50
3S = 1/2 - 1/5 + 1/5 - 1/8 + ... +1/47 - 1/50
3S = 1/2 - 1/50
3S = 12/25
=> S = 12/25 : 3 = 4/25
k, đây là dạng toán sai phân hữu hạn.
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số hạng tổng quát là 1/[n.(n+3)] = (1/3).[(n+3)-n]/[n.(n+3)] = (1/3). [1/n - 1/(n+3)]
=>
A = (1/3).[(1/2 - 1/5) + (1/5 - 1/8) + (1/8 - 1/11) +...+(1/44 - 1/47) + (1/47 - 1/50)]
= (1/3).[1/2 - 1/50]
= (1/3). (24/50) = (1/3).(12/25) = 4/25
vậy A = 4/25
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good luck!