Ta có: \(\hept{\begin{cases}A=\frac{10^{2016}+1}{10^{2017}+1}\\B=\frac{10^{2017}+1}{10^{2018}+1}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}10A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\\10B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\end{cases}}\)
Vì \(\frac{9}{10^{1017}+1}>\frac{9}{10^{2018}+1}\)
nên \(10A>10B\Rightarrow A>B\)
\(A=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\frac{10\cdot(10^{2016}+1)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(A=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{2017}+1\)
\(\Rightarrow\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
\(\Rightarrow\)\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)
....
A= \(\frac{10^{2016}+1}{10^{2016}+1}=\frac{10^{2016}+1}{10\cdot10^{2016}+1}=\frac{1}{10}\cdot\frac{10^{2016}+1}{10^{2016}+1}=\frac{1}{10}\)(1)
B=\(\frac{10^{2017}+1}{10^{2018}+1}=\frac{10^{2017}+1}{10\cdot10^{2017}+1}=\frac{1}{10}\cdot\frac{10^{2017}+1}{10^{2017}+1}=\frac{1}{10}\)(2)
Từ (1) và (2) \(\Rightarrow\)A=B
