Phân tích 2 phân số ta có:
1 = \(\dfrac{2017\times2019}{2017\times2019}\) = \(\dfrac{\left(2018-1\right)\times\left(2018+1\right)}{2017\times2019}\) = \(\dfrac{2018^2-1^2}{2017\times2019}\)
\(\dfrac{2018\times2018}{2017\times2019}\) = \(\dfrac{2018^2}{2017\times2019}\)
Vì \(2018^2\) > \(2018^2-1^2\) nên \(\dfrac{2018^2}{2017\times2019}\) > \(\dfrac{2018^2-1^2}{2017\times2019}\) hay \(\dfrac{2018\times2018}{2017\times2019}\) > 1
(Áp dụng hằng đẳng thức \(a^2-b^2\) = (a - b)(a + b))
\(2017\times2019=\left(2018-1\right)\times\left(2018+1\right)=2018\times2018+2018-2018-1=2018\times2018-1< 2018\times2018\)
\(\Rightarrow\dfrac{2018\times2018}{2017\times2019}>\dfrac{2018\times2018}{2018\times2018}=1\)
