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\(N=\frac{1}{1x5}+\frac{1}{5x10}+...+\frac{1}{2005x2010}\)
\(\Rightarrow5N=\frac{5}{1x5}+\frac{5}{5x10}+\frac{5}{10x15}+...+\frac{5}{2005x2010}\)
\(\Rightarrow5N=1-\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(\Rightarrow5N=1-\frac{1}{5}-\frac{1}{2010}\)
\(\Rightarrow5N=\frac{4}{5}-\frac{1}{2010}\)
\(\Rightarrow5N=\frac{1607}{2010}\)
\(\Rightarrow N=\frac{1607}{10050}\)
Nhấn đúng cho mk nha!!!!!!!!!
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
Ta có:
\(N=\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}...+\frac{1}{2005x2010}\)
\(\Rightarrow Nx5=\left(\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}...+\frac{1}{2005x2010}\right)x5\)
\(=\frac{5}{1x5}+\frac{5}{5x10}+\frac{5}{10x15}...+\frac{5}{2005x2010}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
\(\Rightarrow N=\frac{2009}{2010}:5=\frac{2009}{2010}x\frac{1}{5}=\frac{2009}{10050}\)
Ta có :
\(\frac{1}{2009}+\frac{2}{2009}+....+\frac{2008}{2009}\)
\(=\frac{1+2+....+2008}{2009}\)
\(=\frac{2017036}{2009}=1004\)
Ta có ; \(\frac{1}{2009}+\frac{2}{2009}+\frac{3}{2009}+......+\frac{2008}{2009}\)
\(=\frac{1+2+3+......+2008}{2009}\)
\(=\frac{2017036}{2009}=1004\)
gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
a) Cho: \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(\Rightarrow3A-A=3-\frac{1}{81}\)
\(\Rightarrow A=\frac{3-\frac{1}{81}}{2}\)
\(A=\frac{121}{81}\)
b) \(37,52+4,7\times2,3-9,8\)
\(=37,52+10,81-9,8\)
\(=38,53\)
Chúc bn học tốt !!!!!
a, \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{5}\)+ \(\frac{1}{6}\)
= (\(\frac{1}{2}+\)\(\frac{1}{3}+\)\(\frac{1}{6}\)) + \(\frac{1}{5}\)
= 1 + \(\frac{1}{5}\)
= \(\frac{6}{5}\)
b, mk chịu
a) \(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\frac{1}{5}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)+\frac{1}{5}\)
\(=1+\frac{1}{5}\)
\(=\frac{6}{5}\)
b) \(\frac{4}{6}+\frac{7}{13}+\frac{17}{9}+\frac{19}{13}-\frac{8}{9}+\frac{14}{6}=\left(\frac{4}{6}+\frac{14}{6}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)+\left(\frac{17}{9}-\frac{8}{9}\right)\)
\(=\frac{18}{6}+\frac{26}{13}+\frac{9}{9}=3+2+1=6\)
Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)
\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)
= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + ( \(2\frac{5}{7}\)+ \(4\frac{5}{7}\))
= \(4\frac{4}{5}\) + \(6\frac{5}{7}\)
= \(\frac{24}{5}\) + \(\frac{47}{7}\)
= ...... ( tính nốt nhé )
\(\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+......+\frac{1}{2005.2010}\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+.......+\frac{5}{2005.2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+......+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{1}{5}-\frac{1}{2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\frac{401}{2010}\)
\(=\frac{1}{5}+\frac{401}{10050}=\frac{2411}{10050}\)
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