NT
2
K
Khách
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HN
25 tháng 4 2018
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+.........+\frac{4}{65.68}\)
\(A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.........+\frac{1}{65.68}\right)\)
\(A=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{65.68}\right)\)
\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-...........-\frac{1}{65}-\frac{1}{68}\right)\)
\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(A=\frac{4}{3}\left(\frac{34}{68}-\frac{1}{68}\right)\)
\(A=\frac{4}{3}\left(\frac{33}{68}\right)\)
\(A=\frac{11}{17}\)
Vậy A = \(\frac{11}{17}\)
Chúc bạn học tốt!
PQ
6
9 tháng 8 2015
S = 4/2.5 + 4/5.8 + 4/8.11 + ... + 4/65.48
S = 4/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/65.68 )
S = 4/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/65 - 1/68 )
S = 4/3 . ( 1/2 - 1/68 )
S = 4/3 . 33/68
S = 11/17
NT
2
10 tháng 5 2019
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{65.68}\)
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)
\(A=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{65}-\frac{1}{68}\right)\)
\(A=\frac{4}{3}.\left[\frac{1}{2}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{65}-\frac{1}{65}\right)-\frac{1}{68}\right]\)
\(A=\frac{4}{3}.\left[\frac{1}{2}-\frac{1}{68}\right]\)
\(A=\frac{4}{3}.\frac{33}{68}\)
\(A=\frac{11}{17}\)
~ Hok tốt ~
TN
10 tháng 5 2019
\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)
\(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(=\frac{4}{3}\times\frac{33}{68}=\frac{11}{17}\)
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
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)
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
TM
12 tháng 5 2017
A=...
<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)
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
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{10300}=\frac{1}{x}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{100.103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{x}.3=\frac{101}{206}\)
\(\Rightarrow\frac{1}{x}=\frac{101}{618}\)
\(\Rightarrow x=\frac{618}{101}\)
16 tháng 4 2016
Đề sai rồi! Đề đúng nè: A = 1/2.5 + 1/5.8 +.......+ 1/92.95 + 1/95.98
Bài làm : A =.............Ghi lại đề
3A = 3/2.5 + 3/5.8 +........+ 3/95.98
3A = 1/2 - 1/5 + 1/5 - 1/8 +............+ 1/95 - 1/98
3A = 1/2 - 1/98
3A = 48/98
A = 16/98 = 8/49
16 tháng 4 2016
3A=\(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\)
3A= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{98}\)
3A=\(\frac{1}{2}-\frac{1}{98}\)
3A=\(\frac{49}{98}-\frac{1}{98}\)
3A=\(\frac{24}{49}\)
A=\(\frac{8}{49}\)
ủng hộ mình nha
\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)
\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}\left(\frac{34}{68}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)