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16 tháng 5 2016

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

16 tháng 5 2016

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 \(10^{2007}+1< 10^{2008}+1=>\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}=>1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}=>10A>10B=>A>B\)

21 tháng 4 2021

Hỏi đáp Toán

12 tháng 12 2015

\(1-A=\frac{10^{2007}-10^{2006}}{10^{2007}+1}=\frac{9.10^{2006}}{10^{2007}+1}=\frac{9.2^{2007}}{10^{2008}+10}\)

\(1-B=\frac{10^{2008}-10^{2007}}{10^{2008}+1}=\frac{9.10^{2007}}{10^{2008}+1}\)

=>1-A< 1-B

=> A > B

10 tháng 9 2016

Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)(\(a;b;m\in\)N*)

Ta có: 

\(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)

\(B< \frac{10^{2007}+10}{10^{2008}+10}\)

\(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)

\(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)

=> \(B< A\)

10 tháng 9 2016

thank you

30 tháng 12 2017

\(10A=\dfrac{10^{2007}+10}{10^{2007}+1}=\dfrac{10^{2007}+1+9}{10^{2007}+1}=1+\dfrac{9}{10^{2007}+1}\left(1\right)\)\(10B=\dfrac{10^{2008}+10}{10^{2008}+1}=\dfrac{10^{2008}+1+9}{10^{2008}+1}=1+\dfrac{9}{10^{2008}+1}\left(2\right)\)Từ (1) và ( 2 ) suy ra A>B

Bài 2: 

a: \(5^{2008}+5^{2007}+5^{2006}\)

\(=5^{2006}\left(5^2+5+1\right)=5^{2006}\cdot31⋮31\)

b: \(8^8+2^{20}\)

\(=2^{24}+2^{20}\)

\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

12 tháng 4 2017

\(Tacó:10A=\frac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)\(10B=\frac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)\(Vì:1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)

\(\Rightarrow10A>10B\)

\(\Rightarrow A>B\)