Điều kiện xác định $x\ge -1$
$\begin{array}{l} \left( {\sqrt {x + 3} - \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\\ \Leftrightarrow \left( {\sqrt {x + 3} - \sqrt {x + 1} } \right)\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\\ \Leftrightarrow \left( {x + 3 - x - 1} \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\left( {\sqrt {x + 3} + \sqrt {x + 1} } \right)\\ \Leftrightarrow {x^2} + \sqrt {{x^2} + 4x + 3} = x\sqrt {x + 3} + x\sqrt {x + 1} \\ \Leftrightarrow {x^2} - x\sqrt {x + 1} + \sqrt {\left( {x + 1} \right)\left( {x + 3} \right)} - x\sqrt {\left( {x + 3} \right)} = 0\\ \Leftrightarrow x\left( {x - \sqrt {x + 1} } \right) + \sqrt {x + 3} \left( {\sqrt {x + 1} - x} \right) = 0\\ \Leftrightarrow \left( {x - \sqrt {x + 1} } \right)\left( {x - \sqrt {x + 3} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \sqrt {x + 1} \\ x = \sqrt {x + 3} \end{array} \right. \Rightarrow \left[ \begin{array}{l} {x^2} - x - 1 = 0\\ {x^2} - x - 3 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{1 \pm \sqrt 5 }}{2}\\ x = \dfrac{{1 \pm \sqrt {13} }}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{{1 + \sqrt 5 }}{2}\\ x = \dfrac{{1 - \sqrt 5 }}{2}\\ x = \dfrac{{1 + \sqrt {13} }}{2} \end{array} \right.\\ \Rightarrow S = \left\{ {\dfrac{{1 + \sqrt 5 }}{2};\dfrac{{1 - \sqrt 5 }}{2};\dfrac{{1 + \sqrt {13} }}{2}} \right\} \end{array}$
