Đáp án:
\(\left[ \begin{array}{l}
m = \dfrac{{8 + \sqrt {10} }}{2}\\
m = \dfrac{{8 - \sqrt {10} }}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\Delta ' \ge 0\\
\to 1 - m + 3 \ge 0\\
\to 2 - m \ge 0\\
\to 2 \ge m\\
Có:{x_1}^3{x_2} + {x_1}{x_2}^3 = - 6\\
\to {x_1}{x_2}\left( {{x_1}^2 + {x_2}^2} \right) = - 6\\
\to {x_1}{x_2}\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) = - 6\\
\to {x_1}{x_2}\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right] = - 6\\
\to \left( {m - 3} \right)\left( {4 - 2.\left( {m - 3} \right)} \right) = - 6\\
\to \left( {m - 3} \right)\left( {4 - 2m + 6} \right) = - 6\\
\to \left( {m - 3} \right)\left( {10 - 2m} \right) = - 6\\
\to 4\left( {m - 3} \right)\left( {5 - m} \right) = - 6\\
\to - {m^2} + 8m - 15 = - \dfrac{3}{2}\\
\to - {m^2} + 8m - \dfrac{{27}}{2} = 0\\
\Delta ' = 16 - \left( { - 1} \right)\left( { - \dfrac{{27}}{2}} \right) = \dfrac{5}{2}\\
\to \left[ \begin{array}{l}
m = \dfrac{{8 + \sqrt {10} }}{2}\\
m = \dfrac{{8 - \sqrt {10} }}{2}
\end{array} \right.
\end{array}\)
