Đáp án:

$m \in \{2;4\}.$

Giải thích các bước giải:

$C_1\\ x^2+6x+6m-m^2=0\\ \Delta'=3^2-6m+m^2=(m-3)^2 \ge 0 \ \forall \ m$

$\Rightarrow$ Phương trình luôn có hai nghiệm

$Vi-et: x_1+x_2=-6(1)\\ x_1x_2=6m-m^2\\ \circledast x_1^3-x_2^3+2x_1^2+12x_1+72=0\\ \Leftrightarrow (x_1-x_2)(x_1^2+x_1x_2+x_2^2)+2x_1(x_1+6)+72=0\\ \Leftrightarrow (x_1-x_2)(x_1^2+2x_1x_2+x_2^2-x_1x_2)+2x_1(x_1-x_1-x_2)+72=0\\ \Leftrightarrow (x_1-x_2)[(x_1+x_2)^2-x_1x_2]-2x_1x_2+72=0\\ \Leftrightarrow (x_1-x_2)[(-6)^2+m^2-6m]-2(6m-m^2)+72=0\\ \Leftrightarrow (x_1-x_2)[m^2-6m+36]-2m^2-12m+72=0\\ \Leftrightarrow (x_1-x_2)[m^2-6m+36]-2(m^2-6m+36)=0\\ \Leftrightarrow (x_1-x_2-2)(m^2-6m+36)=0\\ \Leftrightarrow (x_1-x_2-2)(m^2-6m+9+27)=0\\ \Leftrightarrow (x_1-x_2-2)[(m-3)^2+27]=0\\ \Leftrightarrow x_1-x_2-2=0\\ \Leftrightarrow x_1-x_2=2(2)$

Từ $(1),(2)$ ta có hệ: $\left\{\begin{array}{l} x_1+x_2=-6 \\ x_1-x_2=2\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x_1=-2 \\ x_2=-4\end{array} \right.$

$\circledast 6m-m^2=x_1x_2\\ \Leftrightarrow 6m-m^2=8\\ \Leftrightarrow m^2-6m+8=0\\ \Leftrightarrow \left[\begin{array}{l} m=2 \\ m=4\end{array} \right.\\ C_2\\ x^2+6x+6m-m^2=0\\ \Delta'=3^2-6m+m^2=(m-3)^2 \ge 0 \ \forall \ m$

$\Rightarrow$ Phương trình luôn có hai nghiệm

$\left.\begin{array}{l} Vi-et: x_1+x_2=-6(1)\\x_1x_2=6m-m^2=-m(m-6) \end{array} \right\} \Rightarrow \left[\begin{array}{l} x_1=-m; x_2=m-6 \\ x_1=m-6 , x_2=-m\end{array} \right.\\ TH_1: x_1=-m; x_2=m-6\\ x_1^3-x_2^3+2x_1^2+12x_1+72=0\\ \Leftrightarrow (-m)^3-(m-6)^3+2.(-m)^2+12.(-m)+72=0\\ \Leftrightarrow -2 m^3 + 20 m^2 - 120 m + 288=0\\ \Leftrightarrow -2 m^3+8m^2 + 12 m^2 - 48 m -72m+ 288=0\\ \Leftrightarrow -2 m^2(m-4) + 12 m(m - 4) -72(m-4)=0\\ \Leftrightarrow (m-4)(-2 m^2 + 12m -72)=0\\ \Leftrightarrow (m-4)(m^2-6m+36)=0\\ \Leftrightarrow (m-4)[(m-3)^2+27]=0\\ \Leftrightarrow m=4\\ TH_2: x_1=m-6; x_2=-m\\ x_1^3-x_2^3+2x_1^2+12x_1+72=0\\ \Leftrightarrow (m-6)^3-(-m)^3+2(m-6)^2+12(m-6)+72=0\\ \Leftrightarrow 2 m^3 - 16 m^2 + 96 m - 144=0\\ \Leftrightarrow  m^3 - 8 m^2 + 48 m - 72=0\\ \Leftrightarrow  m^3 - 2 m^2 -6m^2+ 12m +36m - 72=0\\ \Leftrightarrow  m^2(m - 2 ) -6m(m-2) +36(m - 2)=0\\ \Leftrightarrow  (m - 2 ) (m^2-6m+36)=0\\ \Leftrightarrow  (m - 2 )[(m-3)^2+27]=0\\ \Leftrightarrow m=2$

Vậy với $m \in \{2;4\}$ phương trình có hai nghiệm thoả mãn đề.

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