Sửa đề:
Nếu:
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{69^{2015}+1}{69^{2017}+1}< 1\)
\(B< \dfrac{69^{2015}+1+68}{69^{2017}+1+68}\Leftrightarrow B< \dfrac{69^{2015}+69}{69^{2017}+69}\)
\(B< \dfrac{69\left(69^{2014}+1\right)}{69\left(69^{2016}+1\right)}\Leftrightarrow B< \dfrac{69^{2014}+1}{69^{2016}+1}=A\)
\(B< A\)
không sửa đề giải luôn:
\(\dfrac{\dfrac{1}{13^2}(13^{2019}+69)+69-\dfrac{69}{13^2}}{13^{2019}+69} \)
\(\\\Rightarrow A=\dfrac{1}{13^2}+\dfrac{69(1-\dfrac{1}{13^2})}{13^{2019}+69}\)
Tương tự: \(B=\dfrac{1}{13^2}+\dfrac{1-\dfrac{1}{13^2}}{13^{2017}+1}\)
Giả sử: \(\dfrac{69}{13^{2019}+69}<\dfrac{1}{13^{2017}+1}\)
\(\\\Rightarrow 13^{2019}>69.13^{2017}\)
Điều này hiển nhiên đúng do \(13^2>69\)
Vậy \(B>A\)