đặt \(2008=a\)

\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=\sqrt{\left(a+1\right)^2-\frac{2\left(a+1\right).a}{a+1}+\left(\frac{a}{a+1}\right)^2}=\)\(\sqrt{\left(a+1-\frac{a}{a+1}\right)^2}=a+1-\frac{a}{a+1}\)=2008+1- \(\frac{2008}{2009}\)

=> A= 2008+1 = 2009

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 \(2008=a\)

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

Phương trình có vô số nghiệm

Nếu thay \(\sqrt{y-2008}\) bằng \(\sqrt{y+2008}\) thì phương trình có bộ nghiệm duy nhất: \(\left(x;y;z\right)=\left(2010;-2007;3\right)\)

Mời bạn đi nối này http://olm.vn/hoi-dap/question/189394.html