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3 tháng 3 2015
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
NT
14 tháng 8 2018
câu hỏi hay......nhưng tui xin nhường cho các bn khác
Hãy tích đúng cho tui nha
THANKS
NL
2
15 tháng 8 2016
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\)
\(3.A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)
\(3.A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\right)\)
\(\Rightarrow2.A=1-\frac{1}{3^{2008}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2008}}\right):2\)
15 tháng 8 2016
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2008}}\)
\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2008}}\right)\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{2007}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2008}}\right)\)
\(2A=\frac{1}{3^{2007}}-\frac{1}{3}\)
\(A=\frac{\frac{1}{3^{2007}}-\frac{1}{3}}{2}\)
12 tháng 2 2018
Ta có :
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)\)
\(B=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)}=\frac{1}{2009}\)
Vậy \(\frac{A}{B}=\frac{1}{2009}\)
12 tháng 2 2018
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{1007}+\frac{1}{2008}\)
\(B=\frac{2008}{1}+1+\frac{2007}{2}+1+\frac{2006}{3}+1+....+\frac{2}{2007}+1+\frac{1}{2008}+1-2008\)
\(B=\frac{2009}{1}+\frac{2009}{2}+\frac{2009}{3}+.....+\frac{2009}{2007}+\frac{2009}{2008}-\frac{2009.2008}{2009}\)
\(B=2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{2008}-\frac{2008}{2009}\right)\)
Từ đó suy ra \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{1008}+\frac{2008}{2009}\right)}\)
\(=\frac{\frac{1}{2009}}{2009\cdot\left(1+\frac{2008}{2009}\right)}\)
Bí òi
13 tháng 9 2016
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)}\)
\(A=\frac{1}{2011}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\right)\)
\(2A=1-\frac{1}{3^{2008}}\)
\(A=\frac{\left(1-\frac{1}{3^{2008}}\right)}{2}\)