Question

cho \(A=\sqrt{2013^2+2013^2.2014^2+2014^2}\)

chứng minh A là 1 số tự nhiên

Answer 1

Ta có :

\(A=\sqrt{2013^2+2013^2.2014^2+2014^2}\)

\(=\sqrt{\left(2013.2014\right)^2+2013.\left(2014-1\right)+\left(2013+1\right).2014}\)

\(=\sqrt{\left(2013.2014\right)^2+2013.2014-2013+2014+2014.2013}\)

\(=\sqrt{\left(2013.2014\right)^2+2.2013.2014.1+1^2}\)

\(=\sqrt{\left(2013.2014+1\right)^2}\)

\(=2013.2014+1\in N\)

Vậy ...

Answer 2

Ta có: \(A=\sqrt{2013^2+2013^2.2014^2+2014^2}\)

<=>\(A=\sqrt{\left(2014^2+2013^2-2.2013.3014\right)+2.2013.2014+\left(2013.2014\right)^2}\)

<=>\(A=\sqrt{\left(2014-2013\right)^2+2.2013.2014+\left(2013.2014\right)^2}\)

<=>\(A=\sqrt{1+2.2013.2014+\left(2013.2014\right)^2}\)

<=>\(A=\sqrt{\left(2013.2014+1\right)^2}\)

<=>A=2013.2014+1

<=>A=4054183

Vậy A là số tự nhiên