Điều kiện xác định $x\ge -1$

$\begin{array}{l} \left( {1 + \sqrt {{x^2} + 2017x + 2016} } \right)\left( {\sqrt {2016 + x}  - \sqrt {x + 1} } \right) = 2015\\  \Leftrightarrow \left( {1 + \sqrt {\left( {x + 1} \right)\left( {x + 2016} \right)} } \right)\dfrac{{2016 + x - x - 1}}{{\sqrt {2016 + x}  + \sqrt {x + 1} }} = 2015\\  \Leftrightarrow \left( {1 + \sqrt {\left( {x + 1} \right)\left( {x + 2016} \right)} } \right)\dfrac{{2015}}{{\sqrt {2016 + x}  + \sqrt {x + 1} }} = 2015\\  \Leftrightarrow 1 + \sqrt {\left( {x + 1} \right)\left( {x + 2016} \right)}  = \sqrt {2016 + x}  + \sqrt {x + 1} \\  \Leftrightarrow \sqrt {\left( {x + 1} \right)\left( {x + 2016} \right)}  - \sqrt {2016 + x}  + 1 - \sqrt {x + 1}  = 0\\  \Leftrightarrow \sqrt {2016 + x} \left( {\sqrt {x + 1}  - 1} \right) - \left( {\sqrt {x + 1}  - 1} \right) = 0\\  \Leftrightarrow \left( {\sqrt {x + 1}  - 1} \right)\left( {\sqrt {2016 + x}  - 1} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} \sqrt {x + 1}  = 1\\ \sqrt {2016 + x}  = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x + 1 = 1\\ 2016 + x = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0(TM)\\ x =  - 2015(L) \end{array} \right.\\  \Rightarrow S = \left\{ 0 \right\} \end{array}$

$ĐKXĐ: x\ge-1$

$(1+\sqrt{x^2+2017x+2016}(\sqrt{2016+x}-\sqrt{x+1})=2015$

$\Leftrightarrow(1+\sqrt{(x+1)(x+2016)}).\dfrac{(\sqrt{2016+x})^2-(\sqrt{x+1})^2}{\sqrt{2016+x}+\sqrt{x+1}}=2015$

$\Leftrightarrow(1+\sqrt{(x+1)(x+2016)}).\dfrac{(2016+x)-(x+1)}{\sqrt{2016+x}+\sqrt{x+1}}=2015$

$\Leftrightarrow(1+\sqrt{(x+1)(x+2016)}).\dfrac{(2016+x)-(x+1)}{\sqrt{2016+x}+\sqrt{x+1}}=2015$

$\Leftrightarrow(1+\sqrt{(x+1)(x+2016)}).\dfrac{2016+x-x-1}{\sqrt{2016+x}+\sqrt{x+1}}=2015$

$\Leftrightarrow(1+\sqrt{(x+1)(x+2016)}).\dfrac{2015}{\sqrt{2016+x}+\sqrt{x+1}}=2015$

$\Leftrightarrow1+\sqrt{(x+1)(x+2016)}=\sqrt{2016+x}+\sqrt{x+1}$

$\Leftrightarrow1+\sqrt{(x+1)(x+2016)}=\sqrt{2016+x}+\sqrt{x+1}$

$\Leftrightarrow1+\sqrt{(x+1)(x+2016)}-\sqrt{2016+x}-\sqrt{x+1}=0$

$\Leftrightarrow(1-\sqrt{x+1})+(\sqrt{(x+1)(x+2016)}-\sqrt{2016+x})=0$

$\Leftrightarrow(1-\sqrt{x+1})+\sqrt{2016+x}(\sqrt{x+1}-1)=0$

$\Leftrightarrow(1-\sqrt{x+1})-\sqrt{2016+x}(1-\sqrt{x+1})=0$

$\Leftrightarrow(1-\sqrt{2016+x})(1-\sqrt{x+1})=0$

$\Leftrightarrow\left[\begin{matrix} 1-\sqrt{2016+x}=0\\ 1-\sqrt{x+1}=0\end{matrix}\right.$

$\Leftrightarrow\left[\begin{matrix} \sqrt{2016+x}=1\\ \sqrt{x+1}=1\end{matrix}\right.$

$\Leftrightarrow\left[\begin{matrix} 2016+x=1\\ x+1=1\end{matrix}\right.$

$\Leftrightarrow\left[\begin{matrix} x=-2015(L)\\ x=0(TM)\end{matrix}\right.$

Vậy $x=0$