\(B=\sqrt{1+2008^2+\dfrac{2008^2}{2009^2}}+\dfrac{2008}{2009}=\sqrt{\dfrac{2009^2+2008^2.2009^2+2008^2}{2009^2}}+\dfrac{2008}{2009}=\dfrac{\sqrt{2009^2+\left(2009-1\right)^2.2009^2+2008^2}}{2009}+\dfrac{2008}{2009}=\dfrac{\sqrt{2009^2+2009^4-2.2009.2009^2+2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4+2.2009^2-2.\left(2008+1\right).2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4+2.2009^2-2.2008.2009^2-2.2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{2009^4-2.2008.2009^2+2008^2}+2008}{2009}=\dfrac{\sqrt{\left(2009^2-2008\right)^2}+2008}{2009}=\dfrac{2009^2-2008+2008}{2009}=2009\in N\)
Vậy B có giá trị là một số tự nhiên
Xét các số thực a, b, c thỏa mãn \(a+b+c=0\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2.\frac{a+b+c}{abc}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
Ta có:
\(B=\sqrt{1+2008^2+\frac{2008^2}{2009^2}}+\frac{2008}{2009}\)
\(=\sqrt{2008^2}.\sqrt{\frac{1}{2018^2}+\frac{1}{1^2}+\frac{1}{2009^2}}+\frac{2008}{2009}\)
\(=2008.\sqrt{\frac{1}{2018^2}+\frac{1}{1^2}+\frac{1}{\left(-2009\right)^2}}+\frac{2008}{2009}\)
\(=2008.\left|\frac{1}{2008}+1-\frac{1}{2009}\right|+\frac{2008}{2009}\)
\(=2008.\left(\frac{1}{2008}+1-\frac{1}{2009}\right)+\frac{2008}{2009}\)
\(=2008.\left(\frac{1}{2008}+1-\frac{1}{2009}+\frac{1}{2009}\right)\)
\(=2008.\frac{2009}{2008}=2009\in\text{N}\)
