Đáp án:
b) \(\dfrac{1}{2} \le m \le 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 3\\
Pt \to {3^2} - 6\left( {m + 1} \right) + {m^2} + 2 = 0\\
\to {m^2} - 6m + 5 = 0\\
\to \left( {m - 1} \right)\left( {m - 5} \right) = 0\\
\to \left[ \begin{array}{l}
m = 1\\
m = 5
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
m = 1\\
m = 5
\end{array} \right.\\
Pt \to \left[ \begin{array}{l}
{x^2} - 4x + 3 = 0\\
{x^2} - 12x + 27 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 3} \right) = 0\\
\left( {x - 3} \right)\left( {x - 9} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 3\\
x = 9
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to {m^2} + 2m + 1 - {m^2} - 2 \ge 0\\
\to 2m - 1 \ge 0\\
\to m \ge \dfrac{1}{2}\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = {m^2} + 2
\end{array} \right.\\
Có:{x_1}^2 + 2\left( {m + 1} \right){x_2} \le 3{m^2} + 10\\
\to {x_1}^2 + \left( {{x_1} + {x_2}} \right){x_2} \le 3{m^2} + 10\\
\to {x_1}^2 + {x_1}{x_2} + {x_2}^2 \le 3{m^2} + 10\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - {x_1}{x_2} \le 3{m^2} + 10\\
\to {\left( {{x_1} + {x_2}} \right)^2} - {x_1}{x_2} \le 3{m^2} + 10\\
\to {\left( {2m + 2} \right)^2} - {m^2} - 2 \le 3{m^2} + 10\\
\to 4{m^2} + 8m + 4 - {m^2} - 2 \le 3{m^2} + 10\\
\to 8m \le 8\\
\to m \le 1\\
KL:\dfrac{1}{2} \le m \le 1
\end{array}\)
