Đáp án:
\[B = 2009\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \sqrt {1 + {{2008}^2} + \dfrac{{{{2008}^2}}}{{{{2009}^2}}}} + \dfrac{{2008}}{{2009}}\\
= \sqrt {\left( {{1^2} + 2.1.2008 + {{2008}^2}} \right) + \dfrac{{{{2008}^2}}}{{{{2009}^2}}} - 2.1.2008} + \dfrac{{2008}}{{2009}}\\
= \sqrt {{{\left( {1 + 2008} \right)}^2} - 2.2008 + \dfrac{{{{2008}^2}}}{{{{2009}^2}}}} + \dfrac{{2008}}{{2009}}\\
= \sqrt {{{2009}^2} - 2.2009.\dfrac{{2008}}{{2009}} + {{\left( {\dfrac{{2008}}{{2009}}} \right)}^2}} + \dfrac{{2008}}{{2009}}\\
= \sqrt {{{\left( {2009 - \dfrac{{2008}}{{2009}}} \right)}^2}} + \dfrac{{2008}}{{2009}}\\
= 2009 - \dfrac{{2008}}{{2009}} + \dfrac{{2008}}{{2009}}\\
= 2009
\end{array}\)
