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a) \(A=1-\frac{1}{2008.2009}\) ; \(B=1-\frac{1}{2009.2010}\)
Vì \(\frac{1}{2008.2009}>\frac{1}{2009.2010}\) nên A < B
Ta có
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\) nên
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)
\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)
Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)
\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(F=\frac{1}{5}-\frac{1}{2009}\)
\(F=\frac{2004}{10045}\)
\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{1}{11.14}+...+\frac{3}{2006.2009}\)
\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(F=\frac{1}{5}-\frac{1}{2009}\)
\(F=0\)
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(=\frac{200}{2009}\)
Gọi \(B=\frac{9}{19}+A\)
\(A=\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+...+\frac{9}{1999\cdot2009}\)
\(\frac{A}{9}=\frac{1}{19\cdot29}+\frac{1}{29\cdot39}+...+\frac{1}{1999\cdot2009}\)
\(\frac{A\cdot10}{9}=\frac{10}{19+29}+\frac{10}{29\cdot39}+...+\frac{10}{1999\cdot2009}\)
\(\frac{A\cdot10}{9}=\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\)
\(\frac{A\cdot10}{9}=\frac{1}{19}-\frac{1}{2009}\)
\(A=\frac{1791}{38171}\)
\(\Rightarrow B=\frac{1}{19}+\frac{1791}{38171}\)
\(\Rightarrow B=\frac{200}{2009}\)
Ta có: \(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
\(\Rightarrow A=\frac{1}{19}+\frac{1791}{38171}\)
\(\Rightarrow A=\frac{200}{2009}\)
Vậy \(A=\frac{200}{2009}.\)
= 5/1-2-3+8/2-3-4+11/3-4-5+...+6026/2008-2009-2010
=3.(5/1-6026/2010)
3.2012/1005
=2012/335
Đặt \(x=2009\)
\(A=2009^8-2010\cdot2009^7+2010\cdot2009^6-2010\cdot2009^5+...+2010\cdot2009^0\)
\(\Leftrightarrow A=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^0\\ \Leftrightarrow A=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...-x^2-x^1+x^1+x^0\)
\(\Leftrightarrow A=x^0\\ \Leftrightarrow A=1\)