Lời giải

Điều kiện xác định x ≥ –2

\(\sqrt {3{\rm{x}} + 15} - \sqrt {4{\rm{x}} + 17} = \sqrt {x + 2} \)

\( \Leftrightarrow \left( {3{\rm{x}} + 15} \right) - \left( {4{\rm{x}} + 17} \right) = \left( {\sqrt {3{\rm{x}} + 15} + \sqrt {4{\rm{x}} + 17} } \right)\sqrt {x + 2} \)

\( \Leftrightarrow - \left( {x + 2} \right) = \left( {\sqrt {3{\rm{x}} + 15} + \sqrt {4{\rm{x}} + 17} } \right)\sqrt {x + 2} \)

\( \Leftrightarrow \left( {x + 2} \right) + \left( {\sqrt {3{\rm{x}} + 15} + \sqrt {4{\rm{x}} + 17} } \right)\sqrt {x + 2} = 0\)

\( \Leftrightarrow \sqrt {x + 2} \left( {\sqrt {x + 2} + \sqrt {3{\rm{x}} + 15} + \sqrt {4{\rm{x}} + 17} } \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\sqrt {x + 2} = 0\\\sqrt {x + 2} + \sqrt {3{\rm{x}} + 15} + \sqrt {4{\rm{x}} + 17} = 0\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\sqrt {x + 2} = 0\\\left\{ \begin{array}{l}\sqrt {x + 2} = 0\\\sqrt {3{\rm{x}} + 15} = 0\\\sqrt {4x + 17} = 0\end{array} \right.\end{array} \right.\)

⇔ x + 2 = 0

⇔ x = –2 (thỏa mãn)

Vậy x = –2.