Đáp án:
\[\sqrt {{{2017}^2} - 1} - \sqrt {{{2016}^2} - 1} > \dfrac{{2.2016}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt {{{2017}^2} - 1} - \sqrt {{{2016}^2} - 1} \\
= \dfrac{{\left( {\sqrt {{{2017}^2} - 1} - \sqrt {{{2016}^2} - 1} } \right)\left( {\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} } \right)}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
= \dfrac{{{{\sqrt {{{2017}^2} - 1} }^2} - {{\sqrt {{{2016}^2} - 1} }^2}}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
= \dfrac{{\left( {{{2017}^2} - 1} \right) - \left( {{{2016}^2} - 1} \right)}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
= \dfrac{{{{2017}^2} - {{2016}^2}}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
= \dfrac{{\left( {2017 - 2016} \right)\left( {2017 + 2016} \right)}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
= \dfrac{{2017 + 2016}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
> \dfrac{{2016 + 2016}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }} = \dfrac{{2.2016}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}\\
\Rightarrow \sqrt {{{2017}^2} - 1} - \sqrt {{{2016}^2} - 1} > \dfrac{{2.2016}}{{\sqrt {{{2017}^2} - 1} + \sqrt {{{2016}^2} - 1} }}
\end{array}\)
