Trước hết ta có:
\(23^{2018}+23^{2020}>2\sqrt{23^{2018}.23^{2020}}=2\sqrt{23^{4038}}=2.23^{2019}\)
Dễ dàng nhận ra \(A>0\) và \(B>0;\) xét thương:
\(\dfrac{A}{B}=\dfrac{23^{2018}+1}{23^{2019}+1}\div\dfrac{23^{2019}+1}{23^{2020}+1}=\dfrac{\left(23^{2018}+1\right)\left(23^{2020}+1\right)}{\left(23^{2019}+1\right)^2}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{23^{4038}+23^{2018}+23^{2020}+1}{\left(23^{2019}+1\right)^2}=\dfrac{\left(23^{2019}\right)^2+23^{2018}+23^{2020}+1}{\left(23^{2019}+1\right)^2}\)
\(\Rightarrow\dfrac{A}{B}>\dfrac{\left(23^{2019}\right)^2+2.23^{2019}+1}{\left(23^{2019}+1\right)^2}=\dfrac{\left(23^{2019}+1\right)^2}{\left(23^{2019}+1\right)^2}=1\)
\(\Rightarrow\dfrac{A}{B}>1\Rightarrow A>B\)
