Giải thích các bước giải:
Đặt $P = \sqrt[]{1+2018^2+\dfrac{2018^2}{2019^2}} + \dfrac{2018}{2019}$
$ = \sqrt[]{1+(2019-1)^2+\dfrac{2018^2}{2019^2}}+\dfrac{2018}{2019}$
$ = \sqrt[]{1+2019^2-2.2019+1 + \bigg(\dfrac{2018}{2019}\bigg)^2}+\dfrac{2018}{2019}$
$ = \sqrt[]{2019^2+2-2.(2018+1) + \bigg(\dfrac{2018}{2019}\bigg)^2}+\dfrac{2018}{2019}$
$ = \sqrt[]{2019^2-2.2018+\bigg(\dfrac{2018}{2019}\bigg)^2}+\dfrac{2018}{2019}$
$ = \sqrt[]{\bigg(2019-\dfrac{2018}{2019}\bigg)^2} + \dfrac{2018}{2019}$
$ = 2019-\dfrac{2018}{2019}+\dfrac{2018}{2019}$
$ = 2019$
Vậy $P$ nhận giá trị tự nhiên là $2019$
