\(B=\sqrt{\frac{2019^2}{2019^2}+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{\left(2018+1\right)^2}{2019^2}+\frac{2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+\frac{2018^2+2.2018+2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+2.2018.\frac{1}{2019}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(\frac{1}{2019}+2018\right)^2}+\frac{2018}{2019}\)
\(B=\frac{1}{2019}+2018+\frac{2018}{2019}=2019\) là một số tự nhiên
\(B=\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{1^2+2018^2+\left(-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2.\frac{2018}{2019}+2.\frac{2018^2}{2019}-2.2018}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2\left(\frac{2018+2018.2018-2018.2019}{2019}\right)}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=1+2018-\frac{2018}{2019}+\frac{2018}{2019}=2019\)
Vậy B có giá trị là 1 số tự nhiên.
Akai Haruma Nguyễn Thanh Hằng Nguyễn Thị Ngọc Thơ