A²=2012²+2012²2013²+2013²

A²=(2013-1)²+2013²+2012²2013²

A²=2.2013²-2.2013+1+2012²2013²

A²=2012²2013²+2.2013(2013-1)+1

A²=(2012.2013+1)² \(\Rightarrow\)A=2012.2013+1 la so tu nhien

k biết lm thì nói toạt ra lun đi

minh bik lam ne

đặt a =2012

\(\Rightarrow A=\sqrt{a^2+a^2\left[a+1\right]^2+\left\{a+1\right\}^2}\)

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

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

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

           \(=a^2+a+1\)

           \(=2012^2+2012+1\) là 1 số tự nhiên

\(A^2=2012^2+2012^2.2013^2+2013^2\)

\(A^2=\left(2013-1\right)^2+2013^2+2012^2.2013^2\)

\(A^2=2.2013^2-2.2013+1+2012^2.2013^2\)

\(A^2=2012^2.2013^2+2.2013.\left(2013-1\right)+1\)

\(A^2=\left(2012.2013+1\right)^2\Rightarrow A=2012.2013+1\) là số tự nhiên

\(A^2=2012^2+2012^2.2013^2+2013^2\)

\(A^2=\left(2013-1\right)^2+2013^2+2012^2.2013^2\)

\(A^2=2.1013^2-2.2013+1+2012^2.2013^2\)

\(A^2=2012^2.2013^2+2.2013\left(2013-1\right)+1\)

\(A^2=\left(2012.2013+1\right)^2\)

\(\Rightarrow A=2012.2013+1\) là số tự nhiên

\(A=\sqrt{2012^2+2012^2.2013^2+2013^2}\)

     \(=\sqrt{2012^2+\left(2012.2013\right)^2+2013^2}\)

       \(=2012+2012.2013+2013\)

Vậy A là một số tự nhiên

P/s: Mình nghĩ thế, không chắc!

\(A=\sqrt{2012^2+2012^2.2013^2+2013^2}\)

\(=\sqrt{\left(2013-1\right)^2+2012^2.2013^2+2013^2}\)

\(=\sqrt{2.2013^2-2.2013+1+2012^2.2013^2}\)

\(=\sqrt{2.2013.\left(2013-1\right)+1+2012^2.2013^2}\)

\(=\sqrt{2012^2.2013^2+2.2013.2012+1}=\sqrt{\left(2012.2013+1\right)^2}=2012.2013+1\)

2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)

\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)

+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)

\(\Rightarrow A< \frac{1}{2}\)

1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)

\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)

\(\Rightarrow A< 2\)

Bài 2 tạm thời chưa nghĩ ra :))

Đặt \(\sqrt{2012}=a;\sqrt{2013}=b\)

Theo đề, ta có: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}-\left(a+b\right)\)

\(=\dfrac{a^3+b^3}{ab}-\dfrac{ab\left(a+b\right)}{ab}\)

\(=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)-ab\left(a+b\right)}{ab}\)

\(=\dfrac{\left(a+b\right)^3-4ab\left(a+b\right)}{ab}\)

\(=\dfrac{\left(a+b\right)\left(a-b\right)^2}{ab}>0\)(đpcm)