\(A=\frac{10^{2006}+1}{10^{2007}+1}=\frac{10^{2006}+1}{\left(10^{2006}+1\right).10}=\frac{1}{10}\)

\(B=\frac{10^{2007}+1}{10^{2008}+1}=\frac{10^{2007}+1}{\left(10^{2007}+1\right).10}=\frac{1}{10}\)

Nếu đề bài là so sách thì A = B

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

\(\Rightarrow10A>10B\)

\(\Rightarrow A>B\)

Đặt A=\(\frac{10^{2006}+1}{10^{2007}+1}\);\(B=\frac{10^{2007}+1}{10^{2008}+1}\)

10A=\(\frac{10\left(10^{2006}+1\right)}{10^{2007}+1}\)=\(\frac{10^{2007}+1+9}{10^{2007}+1}\)

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

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

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

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

=>1-A< 1-B

=> A > B

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

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

\(10A=1\frac{9}{10^{2007}+1}\)

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

Vì \(\frac{9}{10^{2007}+1}\) > \(\frac{9}{10^{2008}+1}\) ==> a > b

K NHA