c)
$x\left ( x + 1 \right )\left ( x + 2 \right )\left ( x + 3 \right ) = 120$
$\Leftrightarrow \left [ x\left ( x + 3 \right ) \right ].\left [ \left ( x + 1 \right )\left ( x + 2 \right ) \right ] = 120$
$\Leftrightarrow \left ( x^{2} + 3x \right ).\left ( x^{2} + 3x + 2 \right ) = 120$
Đặt $t = x^{2} + 3x$ ta có:
$t.\left ( t + 2 \right ) = 120$
$\Leftrightarrow t^{2} + 2t - 120 = 0$
$\Leftrightarrow \left[ \begin{array}{l}t = -12\\t = 10\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}x^{2} + 3x = -12\\x^{2} + 3x = 10\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x^{2} + 3x + 12 = 0\\x^{2} + 3x - 10 = 0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x = -5\\x = 2\end{array} \right.$
