N = \(\dfrac{1}{1x5}+\dfrac{1}{5x10}+.......+\dfrac{1}{2005x2010}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
28 tháng 7 2015
\(N=\frac{1}{1x5}+\frac{1}{5x10}+...+\frac{1}{2005x2010}\)
\(\Rightarrow5N=\frac{5}{1x5}+\frac{5}{5x10}+\frac{5}{10x15}+...+\frac{5}{2005x2010}\)
\(\Rightarrow5N=1-\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(\Rightarrow5N=1-\frac{1}{5}-\frac{1}{2010}\)
\(\Rightarrow5N=\frac{4}{5}-\frac{1}{2010}\)
\(\Rightarrow5N=\frac{1607}{2010}\)
\(\Rightarrow N=\frac{1607}{10050}\)
Nhấn đúng cho mk nha!!!!!!!!!
DD
6 tháng 8 2015
\(N=\frac{1}{1.5}+\frac{1}{5.10}+...+\frac{1}{2005.2010}=\frac{1}{5}\left(\frac{5}{1.5}+\frac{5}{5.10}+...+\frac{5}{2005.2010}\right)=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{2010}\)
5 tháng 4 2017
\(\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+......+\frac{1}{2005.2010}\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+.......+\frac{5}{2005.2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+......+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{1}{5}-\frac{1}{2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\frac{401}{2010}\)
\(=\frac{1}{5}+\frac{401}{10050}=\frac{2411}{10050}\)
5 tháng 4 2017
N = (1/1 - 1/5 + 1/5 -1/10 + ... + 1/2005 - 1/2010 ) x 5
N = (1/1 - 1/2010 ) x5
N = 2009/2010 x5
N = 2009/402
10 tháng 10 2021
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
14 tháng 6 2015
N = 1/1x5 + 1/5x10 + 1/10x15 + 1/15x20 + .....+1/2005 x 2010
N = 1 - 1/5 +1/5-1/5+1/10-1/15+1/5-1/20+.....+1/2005-1/2010
N = 1 - 1/2010
N = 2009/2010
14 tháng 6 2015
Ta có:
\(N=\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}...+\frac{1}{2005x2010}\)
\(\Rightarrow Nx5=\left(\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}...+\frac{1}{2005x2010}\right)x5\)
\(=\frac{5}{1x5}+\frac{5}{5x10}+\frac{5}{10x15}...+\frac{5}{2005x2010}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
\(\Rightarrow N=\frac{2009}{2010}:5=\frac{2009}{2010}x\frac{1}{5}=\frac{2009}{10050}\)
1 tháng 11 2018
a/ \(\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=3\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\right)\)
\(=3\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(=3\left(\dfrac{1}{11}-\dfrac{1}{15}\right)\)
\(=\dfrac{4}{55}\)
b/ \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(=\dfrac{2}{3}\)
c/ \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+.....+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
d/ \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.....+\dfrac{3}{100.103}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{1}{2}-\dfrac{1}{103}\)
\(=\dfrac{101}{206}\)
e/ Đặt :
\(A=\dfrac{1}{1.5}+\dfrac{1}{5.10}+....+\dfrac{1}{95.100}\)
\(\Leftrightarrow5A=\dfrac{5}{1.5}+\dfrac{5}{5.10}+....+\dfrac{5}{95.100}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+....+\dfrac{1}{95}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{100}:5=\dfrac{99}{500}\)
Dấu . là dấu nhân nhé <3
NR
1
28 tháng 6 2021
Ta có: \(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot10}+...+\dfrac{1}{2010\cdot2015}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{2010\cdot2015}\right)\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{2010}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{402}{2015}\)
\(=\dfrac{1}{5}\left(1+\dfrac{402}{2015}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{2417}{2015}=\dfrac{2417}{10075}\)
Lời giải:
$5\times N=1+\frac{5}{5\times 10}+\frac{5}{10\times 15}+....+\frac{5}{2005\times 2010}$
$=1+\frac{10-5}{5\times 10}+\frac{15-10}{10\times 15}+...+\frac{2010-2005}{2005\times 2010}$
$=1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}$
$=1+\frac{1}{5}-\frac{1}{2010}$
$=\frac{2411}{2010}$
$N=\frac{2411}{2010}:5=\frac{2411}{10050}$
\(N=\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{2005.2010}\)
\(\Rightarrow5N=1+\dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+...+\dfrac{5}{2005.2010}\)
\(\Rightarrow5N=1+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+...+\dfrac{1}{2005}-\dfrac{1}{2010}\)
\(\Rightarrow5N=1+\dfrac{1}{5}-\dfrac{1}{2010}\)
\(\Rightarrow5N=\dfrac{2411}{2010}\)
\(\Rightarrow N=\dfrac{2411}{10050}\)