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

Sửa đề$\left( {\sqrt {x + 3}  – \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x$

Điều kiện xác định $x\ge -1$

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