$\begin{array}{l} \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 1} \right)\left( {x + 2} \right) = 60\\ \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 1} \right) = 60\\ \Leftrightarrow \left( {{x^2} - x - 6} \right)\left( {{x^2} - x - 2} \right) = 60\\ \Leftrightarrow \left( {t - 2} \right)\left( {t + 2} \right) = 60\left( {t = {x^2} - x - 4} \right)\\ \Leftrightarrow {t^2} - 4 - 60 = 0\\ \Leftrightarrow {t^2} = 64\\ \Leftrightarrow \left[ \begin{array}{l} t = 8\\ t = - 8 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} {x^2} - x - 4 = 8\\ {x^2} - x - 4 = - 8 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} - x - 12 = 0\\ {x^2} - x + 4 = 0(PTVN) \end{array} \right.\\ \Rightarrow {x^2} - x - 12 = 0\\ \Leftrightarrow \left( {x + 3} \right)\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 3\\ x = 4 \end{array} \right. \end{array}$
2)
$\begin{array}{l} \left( {{x^2} + 14x + 24} \right)\left( {{x^2} + 11x + 24} \right) = 4{x^2}\\ \Leftrightarrow \left( {t + 14x} \right)\left( {t + 11x} \right) = 4{x^2}\left( {{x^2} + 24 = t} \right)\\ \Leftrightarrow {t^2} + 25xt + 150{x^2} = 0\\ \Leftrightarrow {t^2} + 10xt + 15xt + 150{x^2} = 0\\ \Leftrightarrow t\left( {t + 10x} \right) + 15x\left( {t + 10x} \right) = 0\\ \Leftrightarrow \left( {t + 10x} \right)\left( {t + 15x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = - 10x\\ t = - 15x \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + 24 = - 10x\\ {x^2} + 24 = - 15x \end{array} \right.\left( {x < 0} \right)\\ \Leftrightarrow \left[ \begin{array}{l} {x^2} + 10x + 24 = 0\\ {x^2} + 15x + 24 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \left( {x + 6} \right)\left( {x + 4} \right) = 0\\ x = \dfrac{{ - 15 + \sqrt {129} }}{2}\\ x = \dfrac{{ - 15 - \sqrt {129} }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - 4\\ x = - 6\\ x = \dfrac{{ - 15 + \sqrt {129} }}{2}\\ x = \dfrac{{ - 15 - \sqrt {129} }}{2} \end{array} \right. \end{array}$
