Một cách khác nhé!

Đặt a=2014, b=2015 => b-a=1

Khi đó: \(Q=\sqrt{a^2+a^2b^2+b^2}=\sqrt{\left(b-a\right)^2+a^2b^2+2ab}=\sqrt{a^2b^2+2ab+1}=\sqrt{\left(ab+1\right)^2}\)

\(=ab+1=2014.2015+1=4058211\)

Đặt \(2014=a\) thì ta có:

\(Q=\sqrt{a^2+a^2.\left(a+1\right)^2+\left(a+1\right)^2}\)

\(=\sqrt{a^4+2a^3+3a^2+2a+1}\)

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

Vậy Q là số nguyên

vậy là số nguyên

ta có: \(A=\sqrt{1+2.2014+2014^2-2.2014+\frac{2014^2}{2015^2}}+\frac{2014}{2015}.\)

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

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

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

Vậy A=2015

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

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

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

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

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

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

vậy A là số tự nhiên