Ta có: \(10^{2018}>10^{2017}\Rightarrow10^{2018}+1>10^{2017}+1\Rightarrow A=\frac{10^{2018}+1}{10^{2017}+1}>1\) (1)

\(10^{2007}< 10^{2008}\Rightarrow10^{2007}+1< 10^{2008}+1\Rightarrow B=\frac{10^{2007}+1}{10^{2008}+1}< 1\) (2)

Từ (1) và (2) => A > B

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

Ta có: 

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

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

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

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

=> \(B< A\)

thank you

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

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

Nếu đề bài là so sách thì A = B

A>b

Cách làm: Bạn tách |B ra rồi so sánh với từng ps ở A, sau đó Kết luận

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

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

=>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