\(\frac{2^{2017}+1}{2^{2014}+1}>1\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+\left(1+3\right)}{2^{2014}+\left(1+3\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2017}+4}{2^{2014}+4}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{4\left(2^{2015}+1\right)}{4\left(2^{2012}+1\right)}\\ \Rightarrow\frac{2^{2017}+1}{2^{2014}+1}>\frac{2^{2015}+1}{2^{2012}+1}\)
Đặt
\(A=\frac{2^{2015}+1}{2^{2012}+1}\) và \(B=\frac{2^{2017}+1}{2^{2014}+1}.\)
Ta có:
\(\frac{1}{8A}=2^{2015}+\frac{1}{2^{2015}}+8=2^{2015}+8-\frac{7}{2^{2015}}+8=1-\frac{7}{2^{2015}}+8.\)
\(\frac{1}{8B}=2^{2017}+\frac{1}{2^{2017}}+8=2^{2017}+8-\frac{7}{2^{2017}}+8=1-\frac{7}{2^{2017}}+8.\)
Vì \(2^{2015}< 2^{2017}.\)
\(\Rightarrow\frac{7}{2^{2015}}>\frac{7}{2^{2017}}.\)
\(\Rightarrow\frac{7}{2^{2015}}+8>\frac{7}{2^{2017}}+8.\)
\(\Rightarrow1-\frac{7}{2^{2015}}+8< 1-\frac{7}{2^{2017}}+8.\)
\(\Rightarrow A< B.\)
Hay \(\frac{2^{2015}+1}{2^{2012}+1}< \frac{2^{2017}+1}{2^{2014}+1}.\)
Chúc bạn học tốt!