Đáp án:
$m \in \left\{ {2;6} \right\}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{x^2} + 6x + 6m - {m^2} = 0\\
\Leftrightarrow {x^2} - {m^2} + 6x + 6m = 0\\
\Leftrightarrow \left( {x + m} \right)\left( {x - m + 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - m\\
x = m - 6
\end{array} \right.
\end{array}$
Khi đó:
$\begin{array}{l}
x_1^3 - x_2^3 + 2x_1^2 + 12{x_1} + 72 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( { - m} \right)^3} - {\left( {m - 6} \right)^3} + 2{\left( { - m} \right)^2} + 12m + 72 = 0\left( {{x_1} = - m;{x_2} = m - 6} \right)\\
{\left( {m - 6} \right)^3} - {\left( { - m} \right)^3} + 2{\left( {m - 6} \right)^2} + 12\left( {m - 6} \right) + 72 = 0\left( {{x_1} = m - 6;{x_2} = - m} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 2{m^3} + 20{m^2} - 96m + 288 = 0\\
2{m^3} - 16{m^2} + 96m - 144 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 2\left( {m - 6} \right)\left( {{m^2} - 4m + 24} \right) = 0\\
2\left( {m - 2} \right)\left( {{m^2} - 6m + 36} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m = 6\\
{m^2} - 4m + 24 = 0\left( {vn} \right)\\
m = 2\\
{m^2} - 6m + 36 = 0\left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m = 6\\
m = 2
\end{array} \right.
\end{array}$
Vậy $m \in \left\{ {2;6} \right\}$
