LT
1
K
Khách
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LT
3
D
18 tháng 6 2019
Kham khảo tại link:
Câu hỏi của Lê Thúy Hằng - Toán lớp 6 | Học trực tuyến
https://h.vn/hoi-dap/question/821353.html
S
18 tháng 6 2019
\(B=1-\frac{1}{2.5}+1-\frac{1}{5.8}+...+1-\frac{1}{50.53}\)
\(B=17-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{50.53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{53}\right)=17-\frac{1}{3}.\frac{51}{106}=17-\frac{17}{106}=17\left(1-\frac{1}{106}\right)=17.\frac{105}{106}=\frac{1785}{106}\)
SK
1
QD
26 tháng 4 2017
Đề hình như bị sai ban ơi sửa lại
\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)
\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{95}\)
\(A=\dfrac{93}{190}\)
\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)
\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)
\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)
\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)
\(3B=2.\dfrac{93}{190}\)
\(3B=\dfrac{93}{95}\)
\(\Rightarrow B=\dfrac{31}{95}\)
HT
3
ST
19 tháng 4 2021
Ta có : B = 4 / 2. 5 + 4 / 5 . 8 + 4 / 8 . 11 + 4 / 11 . 14
= 4 ( 1 / 2 . 5 + 1 / 5 . 8 + 1 / 8 . 11 + 1 / 11 . 14 )
= 4 . [ 1/3 ( 1 /2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/ 11 + 1/11 - 1/14 )]
= 4 / 3 ( 1/ 2 - 1/14 )
= 4/3 . 3/7
= 4 / 7
Vậy B = 4 / 7
4 tháng 8 2020
\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
Vậy \(A=\frac{504}{1009}.\)
\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)
Vậy \(B=\frac{26}{53}.\)
4 tháng 8 2020
Bài làm:
a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(A=\frac{1}{2}-\frac{1}{2018}\)
\(A=\frac{504}{1009}\)
b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(B=\frac{1}{2}-\frac{1}{106}\)
\(B=\frac{26}{53}\)
NM
2
16 tháng 3 2016
1/2.5+1/5.8+1/8.11+...+1/152.155
=1/3(3/2.5+3/5.8+3/8.11+...+3/152.155
=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/152-1/155)
=1/3(1/2-1/155)
=1/3(155/310-2/310)
=1/3.153/310=51/310
Kết quả:51/310
PN
2
15 tháng 8 2016
\(S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)
\(S=\frac{1}{2}-\frac{1}{101}\)
\(S=\frac{99}{202}\)
NT
4
19 tháng 3 2017
3/2.5 + ...+ 3/17 .20
= 3/2 .(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 120)
= 3/2 . (1/2 - 1/20)
= \(\frac{3}{2}\) . \(\frac{9}{20}\) = \(\frac{27}{40}\)
30 tháng 3 2018
a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=1-\frac{1}{25}\)
\(=\frac{24}{25}\)
30 tháng 3 2018
đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)
\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)
\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)
\(2A=\frac{1}{9}-\frac{1}{45}\)
\(2A=\frac{4}{45}\)
\(A=\frac{4}{45}\div2\)
\(A=\frac{2}{45}\)
\(B=\frac{2.5-1}{2.5}+\frac{5.8-1}{5.8}+\frac{8.11-1}{8.11}+...+\frac{50.53-1}{50.53}\)
\(B=1-\frac{1}{2.5}+1-\frac{1}{5.8}+1-\frac{1}{8.11}+...+1-\frac{1}{50.53}\)
\(B=17-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{50.53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{50}-\frac{1}{53}\right)\)
\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{53}\right)=\frac{1785}{106}\)