Ta có :
\(A=\frac{10^{2016}+1}{10^{2015}+1}=\frac{\left(10^{2016}+1\right).10}{\left(10^{2015}+1\right).10}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10^{2017}+10}{10^{2016}+10}\)
Vì \(10^{2017}=10^{2017}\)và \(10>1\)nên \(10^{2017}+10>10^{2017}+1\)( 1 )
Vì \(10^{2016}=10^{2016}\)và \(10>1\)nên \(10^{2016}+10>10^{2016}+1\)( 2 )
Từ ( 1 ) và ( 2 ) , suy ra : \(\frac{10^{2017}+10}{10^{2016}+10}>\frac{10^{2017}+1}{10^{2016}+1}\)
Vậy \(A>B\)
\(B=\frac{10^{2016}+1}{10^{2017}+1}=\frac{10^{2016}+1+9}{10^{2017}+1+9}=\frac{10^{2016}+10}{10^{2017}+10}=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}=\frac{10^{2015}+1}{10^{2016}+1}\)
lm tương tự vs B ta có
\(A=\frac{10^{2015}+1}{10^{2014}+1}\)
suy ra A>B
Ta có: A=\(\frac{10^{2016}+1}{10^{2015}+1}\)
=>\(\frac{1}{A}=\frac{10^{2015}+1}{10^{2016}+1}=\frac{10\left(10^{2015}+1\right)}{10\left(10^{2016}+1\right)}=\frac{10^{2016}+10}{10\left(10^{2016}+1\right)}=\frac{10^{2016}+1+9}{10\left(10^{2016}+1\right)}\)
\(=\frac{1}{10}+\frac{9}{10^{2017}+10}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}\)
=>\(\frac{1}{B}=\frac{10^{2016}+1}{10^{2017}+1}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2017}+1\right)}=\frac{10^{2017}+10}{10\left(10^{2017}+1\right)}\)
\(=\frac{10^{2017}+1+9}{10\left(10^{2017}+1\right)}=\frac{1}{10}+\frac{9}{10^{2018}+10}\)
Vì\(10^{2017}< 10^{2018}=>10^{2017}+10< 10^{2018}+10\)
\(=>\frac{9}{10^{2017}+10}>\frac{9}{10^{2018}+10}=>\frac{1}{10}+\frac{9}{10^{2017}+10}>\frac{1}{10}+\frac{9}{10^{2017}+10}\)
\(=>\frac{1}{A}>\frac{1}{B}=>A< B\)
