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