Đáp án: $x\in \left \{ \dfrac{-7-\sqrt{57}}{2};0;\dfrac{-7+\sqrt{57}}{2} \right \}$
Giải thích các bước giải:
$x(x+1)(x+6)-x.3=5x\\\Leftrightarrow (x^{2}+x)(x+6)-3x=5x\\\Leftrightarrow x^{3}+6x^{2}+x^{2}+6x-3x=5x\\\Leftrightarrow x^{3}+7x^{2}+3x=5x\\\Leftrightarrow x^{3}+7x^{2}+3x-5x=0\\\Leftrightarrow x^{3}+7x^{2}-2x=0\\\Leftrightarrow x(x^{2}+7x-2)=0\\\Leftrightarrow \left[ \begin{array}{l}x=0\\x^{2}+7x-2=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=0\\x=\dfrac{-7±\sqrt{57}}{2}\end{array} \right.$
