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3 tháng 3 2015
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
LT
1
E
23 tháng 10 2015
Đặt S=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\)
5S=\(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2007}}\)
5S-S=\(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2007}}\)-\(\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\right)\)
4S=\(1-\frac{1}{5^{2008}}\)
=> S=\(\frac{1-\frac{1}{5^{2008}}}{4}\)
28 tháng 11 2016
n) Theo bài ra ta có: \(\frac{x+1}{2008}=\frac{502}{x+1}\)
=> (x+1).(x+1) = 2008.502
=> (x+1)2 = 1008016
=> (x+1)2 = 10042
=> x+1 = 1004
=> x = 2004-1
=> x = 2003
Vậy x = 2003
p) Theo bà ra ta có: \(\left|\frac{5}{4}.x-\frac{7}{2}\right|-\left|\frac{5}{8}.x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}.x-\frac{7}{2}\right|=\left|\frac{5}{8}.x+\frac{3}{5}\right|\)
=> \(\frac{5}{4}.x-\frac{7}{2}=\pm\left(\frac{5}{8}.x+\frac{3}{5}\right)\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{4}.x-\frac{7}{2}=\frac{5}{8}.x+\frac{3}{5}\\\frac{5}{4}.x-\frac{7}{2}=\frac{-5}{8}.x-\frac{3}{5}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{4}.x-\frac{5}{8}.x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}.x+\frac{5}{8}.x=\frac{-3}{5}+\frac{7}{2}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{8}.x=\frac{41}{10}\\\frac{15}{8}.x=\frac{29}{10}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}x=\frac{164}{25}\\x=\frac{116}{75}\end{array}\right.\)
Vậy x=\(\frac{164}{25}\) hoặc x=\(\frac{116}{75}\)
28 tháng 11 2016
\(\frac{x+1}{2008}\)=\(\frac{502}{x+1}\)
=>(x+1)2=502.2008=1008016
=>(x+1)=1004 => x=1004-1=1003
Vậy x=1003
15 tháng 1 2016
Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
11 tháng 8 2019
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)
\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1}{2009}\)
5G= 1+1/5+1/5^2+.....+1/5^2007
4G=5G-G=(1+1/5+1/5^2+....+1/5^2007)-(1/5+1/5^2+1/5^3+....+1/5^2008)
= 1 - 1/5^2008
=>G=(1-1/5^2008)/4
\(G=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2008}}\)(1)
\(\Rightarrow5G=1+\frac{1}{5}+...+\frac{1}{5^{2007}}\)(2)
Lấy (2) trừ đi (1) ta có :
\(4G=1-\frac{1}{5^{2008}}\)
\(\Rightarrow G=\frac{\left(1-\frac{1}{5^{2008}}\right)}{4}\)