Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{1 + \sqrt 5 }}{2}\\
x = \frac{{1 + \sqrt {13} }}{2}
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge - 1\)
Ta có
\(\begin{array}{l}
\left( {\sqrt {x + 3} - \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\,\,\,\, \Rightarrow x > 0\\
\Leftrightarrow \sqrt {x + 3} .\sqrt {{x^2} + 4x + 3} - \sqrt {x + 2} .\sqrt {{x^2} + 4x + 3} + {x^2}\left( {\sqrt {x + 3} - \sqrt {x + 1} } \right) - 2x = 0\\
\Leftrightarrow \left( {x + 3} \right)\sqrt {x + 1} - \left( {x + 1} \right)\sqrt {x + 3} + {x^2}\left( {\sqrt {x + 3} - \sqrt {x + 1} } \right) - 2x = 0\\
\Leftrightarrow \left( {x + 1} \right)\sqrt {x + 1} + 2\sqrt {x + 1} - \left( {x + 1} \right)\sqrt {x + 3} + {x^2}\left( {\sqrt {x + 3} - \sqrt {x + 1} } \right) - 2x = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {\sqrt {x + 1} - \sqrt {x + 3} } \right) + 2\sqrt {x + 1} + {x^2}\left( {\sqrt {x + 3} - \sqrt {x + 1} } \right) - 2x = 0\\
\Leftrightarrow \left( {\sqrt {x + 1} - \sqrt {x + 3} } \right)\left( {x + 1 - {x^2}} \right) + 2\left( {\sqrt {x + 1} - x} \right) = 0\\
\Leftrightarrow \left( {\sqrt {x + 1} - \sqrt {x + 3} } \right)\left( {x + 1 - {x^2}} \right) + 2.\frac{{x + 1 - {x^2}}}{{\sqrt {x + 1} + x}} = 0\\
\Leftrightarrow \left( {x + 1 - {x^2}} \right)\left( {\sqrt {x + 1} - \sqrt {x + 3} + \frac{2}{{\sqrt {x + 1} + x}}} \right) = 0\\
\Leftrightarrow \left( {x + 1 - {x^2}} \right)\left( {\frac{{ - 2}}{{\sqrt {x + 1} + \sqrt {x + 3} }} + \frac{2}{{\sqrt {x + 1} + x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - x - 1 = 0\\
\sqrt {x + 1} + \sqrt {x + 3} = \sqrt {x + 1} + x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{1 + \sqrt 5 }}{2}\left( {t/m} \right)\\
x = \frac{{1 - \sqrt 5 }}{2}\left( L \right)\\
x = \sqrt {x + 3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{1 + \sqrt 5 }}{2}\\
x = \frac{{1 + \sqrt {13} }}{2}
\end{array} \right.
\end{array}\)
