Ta có công thức: \(1+n^2+\frac{n^2}{\left(n+1\right)^2}=\left(1+n-\frac{n}{n+1}\right)^2\)

Chứng minh công thức: \(\left(1+n-\frac{n}{n+1}\right)^2\)

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

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

\(=1+n^2+\frac{n^2-2n^3-2n^2+2n\left(n^2+2n+1\right)-2n^2-2n}{\left(n+1\right)^2}\)

\(=1+n^2+\frac{-2n^3-3n^2-2n+2n^3+4n^2+2n}{\left(n+1\right)^2}=1+n^2+\frac{n^2}{\left(n+1\right)^2}\)

\(A=\sqrt{1+2008^2+\frac{2008^2}{2009^2}}+\frac{2008}{2009}\)

\(=1+2008-\frac{2008}{2009}+\frac{2008}{2009}\)

=2009

=>A là số tự nhiên

Bài 1: 

Ta có: \(a+b\ge2\sqrt{ab}\)

\(b+c\ge2\sqrt{bc}\)

\(a+c\ge2\sqrt{ac}\)

Do đó: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

hay \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)

`A=\sqrt{1+2008^2+2008^2/2009^2}+2008/2009`

`=\sqrt{1+2008^2+2.2008+2008^2/2009^2-2.2008}+2008/2009`

`=\sqrt{(2008+1)^2-2.2008+2008^2/2009^2}+2008/2009`

`=\sqrt{2009-2.2008/2009*2009+2008^2/2009^2}+2008/2009`

`=\sqrt{(2009-2008/2009)^2}+2008/2009`

`=|2009-2008/2009|+2008/2009`

`=2009-2008/2009+2008/2009`

`=2009` là 1 số tự nhiên

Đặt \(2008=a\)

\(\Leftrightarrow A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-\dfrac{2a\left(a+1\right)}{a+1}+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2009\left(đpcm\right)\)

Ta có :\(2009^2=\left(1+2008\right)^2=1+2008^2+2.2008\)

\(\Rightarrow1+2008^2=2009^2-2.2008\)

\(\Rightarrow A=\sqrt{2009^2-2.2008+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}=\sqrt{\left(2009-\dfrac{2008}{2009}\right)^2}+\dfrac{2008}{2009}\)

\(=2009-\dfrac{2008}{2009}+\dfrac{2008}{2009}=2009\)
Vậy A là 1 số tự nhiên.

Sorry thiếu với \(\forall m\inℝ\)

với cả  : P(x) = ax² + bx +c , a khác 0

Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}=\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)

Vì \(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\) nên \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)

\(\Rightarrow\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)>\sqrt{2009}+\sqrt{2008}\)

Hay \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}>\sqrt{2008}+\sqrt{2009}\)

Cảm ơn bạn CTV 

so sánh nha mn