Question

Chứng minh A = \(\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)          là  số tự nhiên                                                                         

Answer 1

A = \(\sqrt{1+2014^2+\frac{2014^2}{2015^2}}\)+ 2014/2015

= \(\sqrt{\frac{2015^2+2014^2.2015^2+2014^2}{2015^2}}\)+ 2014/2015

=\(\frac{\sqrt{2015^2+2014^2.2015^2+2014^2}}{2015}\)+ 2014/2015

Xét 2015^2 + 2014^2.2015^2 + 2014^2

= 2014.2015 + 2015 + 2014^2.2015^2 + 2014.2015 - 2014

= 2014^2.2015^2 + 2.2014.2015 + 1 = (2014.2015 + 1)^2

=> A = \(\frac{2014.2015+1}{2015}\)+ 2014/2015 = \(\frac{2014.2015+2015+1}{2015}\)

 = \(\frac{2014.2016+1}{2015}\) = \(\frac{2015^2-1+1}{2015}\)= 2015 là số tự nhiên 

=> ĐPCM

Answer 2

\(A=\sqrt{\left(2014+1\right)^2-2.2014.1+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)

\(=\sqrt{2015^2-2.2015.\frac{2014}{2015}+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)

\(=\sqrt{\left(2015-\frac{2014}{2015}\right)^2}+\frac{2014}{2015}\)

\(=\left|2015-\frac{2014}{2015}\right|+\frac{2014}{2015}\)

\(=2015-\frac{2014}{2015}+\frac{2014}{2015}=2015\)

vậy A là số tự nhiên