Question

[Thắng Trịnh]

B= \(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)

chứng minh biểu thức B có giá trị nguyên

Answer 1

[_Guiltykamikk_]

\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)

\(B=\sqrt{\left(1+2.2017+2017^2\right)-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)

\(B=\sqrt{\left(1+2017\right)^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)

\(B=\sqrt{2018^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)

\(B=\sqrt{\left(2018-\frac{2017}{2018}\right)^2}+\frac{2017}{2018}\)

Mà  \(\frac{2017}{2018}< 1\Rightarrow2018-\frac{2017}{2018}>0\)

\(\Rightarrow B=2018-\frac{2017}{2018}+\frac{2017}{2018}\)

\(B=2018\)

Vậy bt B có giá trị nguyên 

Answer 2

[Thắng Trịnh]

Cảm ơn bạn mk vừa đăng lên thì đã thấy luôn cách giải 😂