#)Giải :
Ta có : \(A=\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2019}< 1+1+1\)
\(\Rightarrow A< 3\)
Mình giải thế này cho ngắn gọn, với lại nhanh ^^
A = 2017/2018 + 2018/2019+2019/2017 với 3
= \(\frac{2017}{2018}\)+ \(\frac{2019-1}{2019}\)+ \(\frac{2017+1+1}{2019}\)
= 1 - \(\frac{1}{2018}\)+ 1 - \(\frac{1}{2019}\)+ 1 + \(\frac{1}{2017}\)+ \(\frac{1}{2017}\)
= 3 + ( \(\frac{1}{2017}\)- \(\frac{1}{2018}\)) + ( \(\frac{1}{2017}\)- \(\frac{1}{2019}\))
Vì \(\frac{1}{2017}\)> \(\frac{1}{2018}\), \(\frac{1}{2017}\)> \(\frac{1}{2019}\)nên A > 3