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Bài 1:
Ta có: 200920=(20092)10=403608110 ; 2009200910=2009200910
Vì 403608110< 2009200910 => 200920< 2009200910
Bài 1:
Ta có:\(2009^{20}\)=\(2009^{10}\).\(2009^{10}\)
\(20092009^{10}\)=(\(\left(2009.10001\right)^{10}=2009^{10}.10001^{10}\)
Vì 2009<10001\(\Rightarrow2009^{20}< 20092009^{10}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
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 \(\frac{A}{B}=\frac{1}{2009}\)
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 \(\frac{A}{B}=\frac{1}{2009}\)
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{1007}+\frac{1}{2008}\)
\(B=\frac{2008}{1}+1+\frac{2007}{2}+1+\frac{2006}{3}+1+....+\frac{2}{2007}+1+\frac{1}{2008}+1-2008\)
\(B=\frac{2009}{1}+\frac{2009}{2}+\frac{2009}{3}+.....+\frac{2009}{2007}+\frac{2009}{2008}-\frac{2009.2008}{2009}\)
\(B=2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{2008}-\frac{2008}{2009}\right)\)
Từ đó suy ra \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{1008}+\frac{2008}{2009}\right)}\)
\(=\frac{\frac{1}{2009}}{2009\cdot\left(1+\frac{2008}{2009}\right)}\)
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