Đáp án:
\[B = 2018\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \sqrt {1 + {{2017}^2} + \dfrac{{{{2017}^2}}}{{{{2018}^2}}}} + \dfrac{{2017}}{{2018}}\\
= \sqrt {\left( {1 + 2.2017 + {{2017}^2}} \right) - 2.2017 + {{\left( {\dfrac{{2017}}{{2018}}} \right)}^2}} + \dfrac{{2017}}{{2018}}\\
= \sqrt {\left( {{{2017}^2} + 2.2017.1 + {1^2}} \right) - 2.2017 + {{\left( {\dfrac{{2017}}{{2018}}} \right)}^2}} + \dfrac{{2017}}{{2018}}\\
= \sqrt {{{\left( {2017 + 1} \right)}^2} - 2.2017 + {{\left( {\dfrac{{2017}}{{2018}}} \right)}^2}} + \dfrac{{2017}}{{2018}}\\
= \sqrt {{{2018}^2} - 2.2018.\dfrac{{2017}}{{2018}} + \left( {{{\dfrac{{2017}}{{2018}}}^2}} \right)} + \dfrac{{2017}}{{2018}}\\
= \sqrt {{{\left( {2018 - \dfrac{{2017}}{{2018}}} \right)}^2}} + \dfrac{{2017}}{{2018}}\\
= \left| {2018 - \dfrac{{2017}}{{2018}}} \right| + \dfrac{{2017}}{{2018}}\\
= \left( {2018 - \dfrac{{2017}}{{2018}}} \right) + \dfrac{{2017}}{{2018}}\\
= 2018
\end{array}\)
