Ta có: \(\left( {x + 1} \right)\left( {x + 4} \right) - 3\sqrt {{x^2} + 5{\rm{x}} + 2} = 6\)

\( \Leftrightarrow {x^2} + 5{\rm{x}} + 4 - 3\sqrt {{x^2} + 5{\rm{x}} + 2} = 6\)

\( \Leftrightarrow {x^2} + 5{\rm{x}} + 2 - 3\sqrt {{x^2} + 5{\rm{x}} + 2} = 4\)

Đặt \(t = \sqrt {{x^2} + 5{\rm{x}} + 2} \left( {t \ge 0} \right)\)

Suy ra t² – 3t = 4

⇔ t² – 3t – 4 = 0

⇔ t² + t – 4t – 4 = 0

⇔ t(t + 1) – 4(t + 1) = 0

⇔ (t + 1)(t – 4) = 0

\( \Leftrightarrow \left[ \begin{array}{l}t + 1 = 0\\t - 4 = 0\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}t = - 1\\t = 4\end{array} \right.\)

Mà t ≥ 0 nên t = 4

Suy ra \(\sqrt {{x^2} + 5{\rm{x}} + 2} = 4\)

⇔ x² + 5x + 2 = 16

⇔ x² + 5x – 14 = 0

⇔ x² + 7x – 2x – 14 = 0

⇔ x(x + 7) – 2(x + 7) = 0

⇔ (x + 7)(x – 2) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x + 7 = 0\\x - 2 = 0\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = - 7\\x = 2\end{array} \right.\)

Vậy x = 2 hoặc x = –7.