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Đề của bạn sai rồi: Phải là B = \(\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\) chứ ?!
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\(\Rightarrow2009^{20}< 20092009^{10}\)
Gọi \(S=\frac{2009}{1}+\frac{2008}{2}+...+\frac{1}{2009}\)
\(\Rightarrow S=\frac{2010-1}{1}+\frac{2010-2}{2}+...+\frac{2010-2009}{2009}\)
\(\Rightarrow S=2010-1+\frac{2010}{2}-1+...+\frac{2010}{2009}-1\)
\(\Rightarrow S=2010+\frac{2010}{2}+...+\frac{2010}{2009}-\left(1+1+..+1\right)\)
\(\Rightarrow S=2010+\frac{2010}{2}+...+\frac{2010}{2009}-2009\)
\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+...+\frac{2010}{2009}+1\)
\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+..+\frac{2010}{2009}+\frac{2010}{2010}\)
\(\Rightarrow S=2010\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)\)
Khi đó \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}}{2010\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)}=\frac{1}{2010}\)
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)