Đáp án:
\(\left[ \begin{array}{l}
m = \dfrac{5}{4}\\
m = 1
\end{array} \right.\)
Giải thích các bước giải:
c) Để phương trình có nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to 9 - 4m \ge 0\\
\to \dfrac{9}{4} \ge m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = m
\end{array} \right.\\
{x_1}^3{x_2} + {x_1}{x_2}^3 - 2{\left( {{x_1}{x_2}} \right)^2} = 5\\
\to {x_1}{x_2}\left( {{x_1}^2 + {x_2}^2} \right) - 2{\left( {{x_1}{x_2}} \right)^2} = 5\\
\to {x_1}{x_2}\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) - 2{\left( {{x_1}{x_2}} \right)^2} = 5\\
\to {x_1}{x_2}\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right] - 2{\left( {{x_1}{x_2}} \right)^2} = 5\\
\to m\left( {9 - 2m} \right) - 2{m^2} = 5\\
\to - 4{m^2} + 9m - 5 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{5}{4}\\
m = 1
\end{array} \right.\left( {TM} \right)
\end{array}\)
