Đáp án:
\(\left[ \begin{array}{l}
m \ge 4\\
m \le 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 1\\
{x_1}{x_2} = m - 3
\end{array} \right.\\
{x_1}^2 + {x_2} \ge {m^2} - 6m + 8\\
\to {x_1}^2 + \left( {{x_1} + {x_2}} \right){x_2} \ge {m^2} - 6m + 8\\
\to {x_1}^2 + {x_1}{x_2} + {x_2}^2 \ge {m^2} - 6m + 8\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - {x_1}{x_2} \ge {m^2} - 6m + 8\\
\to {\left( {{x_1} + {x_2}} \right)^2} - {x_1}{x_2} \ge {m^2} - 6m + 8\\
\to 1 - m + 3 \ge {m^2} - 6m + 8\\
\to {m^2} - 5m + 4 \ge 0\\
\to \left( {m - 1} \right)\left( {m - 4} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 1 \ge 0\\
m - 4 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 1 \le 0\\
m - 4 \le 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
m \ge 4\\
m \le 1
\end{array} \right.
\end{array}\)
