Giải phương trình: (2x + 1)(2x + 3)(x + 1)² = 18.

\(\begin{array}{l}\left( {2x + 1} \right)\left( {2x + 3} \right){\left( {x + 1} \right)^2} = 18\\ \Leftrightarrow \left( {2x + 2 - 1} \right)\left( {2x + 2 + 1} \right){\left( {x + 1} \right)^2} = 18\\ \Leftrightarrow \left( {{{\left( {2x + 2} \right)}^2} - 1} \right){\left( {x + 1} \right)^2} = 18\\ \Leftrightarrow 4{\left( {x + 1} \right)^4} - {\left( {x + 1} \right)^2} - 18 = 0\end{array}\)

Đặt t = (x + 1)² (t ≥ 0)

Phương trình trở thành:

\[4{t^2} - t - 18 = 0 \Leftrightarrow \left[ \begin{array}{l}t = \frac{9}{4}\\t = - 2(L)\end{array} \right.\]

Với \(t = \frac{9}{4} \Rightarrow {\left( {x + 1} \right)^2} = \frac{9}{4} \Leftrightarrow \left[ \begin{array}{l}x + 1 = \frac{3}{2}\\x + 1 = \frac{{ - 3}}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{1}{2}\\x = \frac{5}{2}\end{array} \right.\).