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

$ĐKXĐ: x\ge-1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Vậy $x=0$