Giả sử A=\(\frac{2^{2015}+1}{2^{2012}+1}\)
-->\(\frac{1}{2^3}A=\frac{2^{2015}+1}{2^{2015}+8}\)
\(\frac{1}{8}A=\frac{2^{2015}+1}{2^{2015}+1}+\frac{2^{2015}+1}{7}\)
\(\frac{1}{8}A=1+\frac{2^{2015}+1}{7}\)
B=\(\frac{2^{2017}+1}{2^{2014}+1}\)
\(\frac{1}{2^3}B=\frac{2^{2017}+1}{2^{2017}+8}\)
\(\frac{1}{8}B=\frac{2^{2017}+1}{2^{2017}+1}+\frac{2^{2017}+1}{7}\)
\(\frac{1}{8}B=1+\frac{2^{2017}+1}{7}\)
Vì \(1+\frac{2^{2015}+1}{7}< 1+\frac{2^{2017}+1}{7}\)
nên \(\frac{1}{8}A< \frac{1}{8}B\)
-->A<B
-->\(\frac{2^{2015}+1}{2^{2012+1}}< \frac{2^{2017+1}}{2^{2014}+1}\)
