Đáp án đúng: B
Giải chi tiết:Ta có: \(\left( x-1 \right)\left( x-3 \right)\left( x+4 \right)\left( x+6 \right)=240\)
\(\begin{align} & \Leftrightarrow \left( \left( x-1 \right)\left( x+4 \right) \right)\left( \left( x-3 \right)\left( x+6 \right) \right)=240 \\ & \Leftrightarrow \left( {{x}^{2}}+3x-4 \right)\left( {{x}^{2}}+3x-18 \right)=240\,\,\,\,\left( * \right) \\ \end{align}\)
Đặt \({{x}^{2}}+3x=t.\)
\(\begin{array}{l} \Rightarrow \left( * \right) \Leftrightarrow \left( {t - 4} \right)\left( {t - 18} \right) = 240\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {t^2} - 22t + 72 - 240 = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {t^2} - 22t - 168 = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left( {t - 28} \right)\left( {t + 6} \right) = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}t - 28 = 0\\t + 6 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}t = 28\\t = - 6\end{array} \right..\end{array}\)
+) Với \(t=28\Rightarrow {{x}^{2}}+3x=28\)
\(\begin{array}{l} \Leftrightarrow {x^2} + 3x - 28 = 0\\ \Leftrightarrow \left( {x - 4} \right)\left( {x + 7} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 4 = 0\\x + 7 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 4\\x = - 7\end{array} \right..\end{array}\)
+) Với \(t=-6\Rightarrow {{x}^{2}}+3x=-6\)
\(\Leftrightarrow {{x}^{2}}+3x+6=0\)
Có \(\Delta ={{3}^{2}}-4.6=9-24=-15<0\)
\(\Rightarrow Pt\) vô nghiệm.
Vậy phương trình có nghiệm \(S=\left\{ 4;-7 \right\}.\)
Chọn B.