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Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)(\(a;b;m\in\)N*)
Ta có:
\(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)
\(B< \frac{10^{2007}+10}{10^{2008}+10}\)
\(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)
\(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)
=> \(B< A\)
\(1-A=\frac{10^{2007}-10^{2006}}{10^{2007}+1}=\frac{9.10^{2006}}{10^{2007}+1}=\frac{9.2^{2007}}{10^{2008}+10}\)
\(1-B=\frac{10^{2008}-10^{2007}}{10^{2008}+1}=\frac{9.10^{2007}}{10^{2008}+1}\)
=>1-A< 1-B
=> A > B
10A=10*\(\frac{10^{2006}+1}{10^{2007}+1}\) 10B=10*\(\frac{10^{2007}+1}{10^{2008}+1}\)
10A=\(\frac{10^{2007}+1+9}{10^{2007}+1}\) 10B=\(\frac{10^{2008}+1+9}{10^{2008}+1}\)
10A=1+\(\frac{9}{10^{2007}+1}\) 10B=1+\(\frac{9}{10^{2008}+1}\)
Vì \(\frac{9}{10^{2007}+1}\)>\(\frac{9}{10^{2008}+1}\)=>1+\(\frac{9}{10^{2007}+1}\)>1+\(\frac{9}{10^{2008}+1}\)
Nên 10A>10B=>A>B
Ta có: \(A=\frac{10^{2006}+1}{10^{2007}+1}\)
\(=>10A=\frac{10^{2007}+10}{10^{2007}+1}=\frac{10^{2007}+1+9}{10^{2007}+1}=\frac{10^{2007}+1}{10^{2007}+1}+\frac{9}{10^{2007}+1}=1+\frac{9}{10^{2007}+1}\)
\(B=\frac{10^{2007}+1}{10^{2008}+1}\)
\(=>10B=\frac{10^{2008}+10}{10^{2008}+1}=\frac{10^{2008}+1+9}{10^{2008}+1}=\frac{10^{2008}+1}{10^{2008}+1}+\frac{9}{10^{2008}+1}=1+\frac{9}{10^{2008}+1}\)
Vì \(10^{2007}+1< 10^{2008}+1=>\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}=>1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}=>10A>10B=>A>B\)
\(Tacó:10A=\frac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)\(10B=\frac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)\(Vì:1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
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\(\Rightarrow2009^{20}< 20092009^{10}\)
\(10A=\dfrac{10^{2007}+10}{10^{2007}+1}=\dfrac{10^{2007}+1+9}{10^{2007}+1}=1+\dfrac{9}{10^{2007}+1}\left(1\right)\)\(10B=\dfrac{10^{2008}+10}{10^{2008}+1}=\dfrac{10^{2008}+1+9}{10^{2008}+1}=1+\dfrac{9}{10^{2008}+1}\left(2\right)\)Từ (1) và ( 2 ) suy ra A>B