Cho $f(x)={{e}^{\sqrt{1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left(...

$1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}=\dfrac{{{x}^{2}}{{\left( x+1 \right)}^{2}}+{{\left( x+1 \right)}^{2}}+{{x}^{2}}}{{{x}^{2}}{{\left( x+1 \right)}^{2}}}=\dfrac{{{\left( x\left( x+1 \right)+1 \right)}^{2}}}{{{x}^{2}}{{\left( x+1 \right)}^{2}}}.$

$f(x)={{e}^{\sqrt{1+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{\left( x+1 \right)}^{2}}}}}}={{e}^{\dfrac{x\left( x+1 \right)+1}{x\left( x+1 \right)}}},\forall x>0$.
Xét dãy số $\left( {{u}_{k}} \right):{{u}_{k}}=\dfrac{k\left( k+1 \right)+1}{k\left( k+1 \right)}=1+\dfrac{1}{k\left( k+1 \right)}=1+\dfrac{1}{k}-\dfrac{1}{k+1}.\left( k\in {{\mathbb{N}}^{*}} \right).$
Ta có ${{u}_{1}}=1+\dfrac{1}{1}-\dfrac{1}{2},{{u}_{2}}=1+\dfrac{1}{2}-\dfrac{1}{3},{{u}_{3}}=1+\dfrac{1}{3}-\dfrac{1}{4},...{{u}_{2017}}=1+\dfrac{1}{2017}-\dfrac{1}{2018}.$
$\begin{aligned}
& f\left( 1 \right).f\left( 2 \right).f\left( 3 \right)...f\left( 2017 \right)={{e}^{{{u}_{1}}+{{u}_{2}}+{{u}_{3}}+...+{{u}_{2017}}}}. \\
& {{u}_{1}}+{{u}_{2}}+{{u}_{3}}+...+{{u}_{2017}}=2017+\dfrac{1}{1}-\dfrac{1}{2018}=\dfrac{{{2018}^{2}}-1}{2018}=\dfrac{m}{n}. \\
\end{aligned}$
Vậy $m-{{n}^{2}}=-1.$