Giải pt  (left( {sqrt {x + 3}  - sqrt {x + 1} } ight)left( {{x^2} + sqrt {{x^2} + 4x + 3} } ight) = 2x)

* Hướng dẫn giải

Đkxđ: \(x \ge  - 1\)

\(\left( {\sqrt {x + 3}  - \sqrt {x + 1} } \right)\left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = 2x\)

\( \Leftrightarrow \left( {{x^2} + \sqrt {{x^2} + 4x + 3} } \right) = \frac{{2x}}{{\left( {\sqrt {x + 3}  - \sqrt {x + 1} } \right)}}\) vì \(\sqrt {x + 3}  - \sqrt {x + 1}  \ne 0\forall x \in \) đkxđ

\( \Leftrightarrow \left( {{x^2} + \sqrt {(x + 3)(x + 1)} } \right) = \frac{{2x\left( {\sqrt {x + 3}  + \sqrt {x + 1} } \right)}}{2}\) vì \(\sqrt {x + 3}  + \sqrt {x + 1}  \ne 0\forall x \in \) đkxđ

\(\begin{array}{l}
 \Leftrightarrow {x^2} + \sqrt {(x + 3)(x + 1)}  - x\sqrt {x + 3}  - x\sqrt {x + 1}  = 0\\
 \Leftrightarrow (x - \sqrt {x + 3} )(x - \sqrt {x + 1} ) = 0 \Leftrightarrow ...x = \frac{{1 + \sqrt {13} }}{2};x = \frac{{1 + \sqrt 5 }}{2}
\end{array}\)