Giải PT ($\sqrt[ ]{x +3}$ – $\sqrt[ ]{x +1}$ ) ($x^{2}$ +$\sqrt[ ]{x^{2} + 4x + 3}$ ) = 2x

Điều kiện xác định $x\ge -1$

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