Ta có:
\(\dfrac{2021}{2018}=\dfrac{2018+3}{2018}=1+\dfrac{3}{2018}\)
\(\dfrac{97968}{97965}=\dfrac{97965+3}{97965}=1+\dfrac{3}{97965}\)
Mà \(2018< 97965\) nên \(\dfrac{3}{2018}>\dfrac{3}{97965}\)
Từ đó suy ra \(1+\dfrac{3}{2018}>1+\dfrac{3}{97965}\) hay\(\dfrac{2021}{2018}>\dfrac{97968}{97965}\)