Question

Chứng minh:

 \(\sqrt{2019^2+2019^2.2020^2+2020^2}\in N\)

Answer 1

Hướng dẫn:

Dat:   \(2019=a\)

Ta có:   \(a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2\)

\(=a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2\)

\(=a^2\left(a^2+2a+2\right)+\left(a+1\right)^2\)

\(=a^4+2a^2\left(a+1\right)+\left(a+1\right)^2\)

\(=\left(a^2+a+1\right)^2\)