Đề bài: Cho  Biết rằng \(f\left( 1 \right).f\left( 1 \right).f\left( 3 \right)…f\left( {2017} \right) = {e^{\frac{m}{n}}}\) với m. n là các số tự nhiên và  tối giản. Tính 

Đặt  \(g\left( x \right) = \sqrt {1 + \frac{1}{{{x^2}}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}} = \frac{{\sqrt {{x^2} + {{\left( {x + 1} \right)}^2} + {x^2}{{\left( {x + 1} \right)}^2}} }}{{x\left( {x + 1} \right)}}\)

\(= \frac{{{x^2} + x + 1}}{{x\left( {x + 1} \right)}}1 + \frac{1}{x} – \frac{1}{{x + 1}}\)

Suy ra \(g\left( 1 \right) + g\left( 2 \right) + g\left( 3 \right) + … + g\left( {2017} \right)\)  

\(= 1 + \frac{1}{2} – \frac{1}{2} + 1 + \frac{1}{2} – \frac{1}{3} + … + 1 + \frac{1}{{2017}} – \frac{1}{{2018}} = 2018 – \frac{1}{{2018}}.\)

Khi đó \(f\left( 1 \right).f\left( 2 \right).f\left( 3 \right)…f\left( {2017} \right) = {e^{g\left( 1 \right) + g\left( 2 \right) + g\left( 3 \right) + … + g\left( {2017} \right)}} = {e^{2018 – \frac{1}{{2018}}}}\) 

\(= {e^{\frac{{{{2018}^2} – 1}}{{2018}}}} = {e^{\frac{m}{n}}} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {m = {{2018}^2} – 1}\\ {n = 2018} \end{array}} \right.\)

Vậy phép tính \(m – {n^2} = {2018^2} – 1 – {2018^2} = – 1\).