Ta có:
\(\frac{2017.2019}{2018.2018}\)
\(=\frac{2017.\left(2018+1\right)}{\left(2017+1\right).2018}\)
\(=\frac{2017.2018+2017}{2017.2018+2018}\)
Vì \(2017.2018+2017< 2017.2018+2018\)( tử nhỏ hơn mẫu )
\(\Rightarrow\frac{2017.2018+2017}{2017.2018+2018}< 1\)
Vậy \(\frac{2017.2019}{2018.2018}< 1\)
( Mk nghĩ vậy )
~~~~~~~Hok tốt~~~~~~~
\(\frac{2017.2019}{2018.2018}=\frac{2017.\left(2018+1\right)}{2018.\left(2017+1\right)}=\frac{2017.2018+2017}{2018.2017+2018}\)
\(2017< 2018\Rightarrow2017.2018+2017< 2018.2017+2018\Rightarrow\frac{2017.2018+2017}{2018.2017+2018}< 1\Rightarrow\frac{2017.2019}{2018.2018}< 1\)