Đáp án:
b) \(\left[ \begin{array}{l}
m = 1\\
m = - 3
\end{array} \right.\)
Giải thích các bước giải:
a)Thay m=-2
\(\begin{array}{l}
Pt \to {x^2} - 2x - 3 = 0\\
\to \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 1 - m + 1 \ge 0\\
\to 2 \ge m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
{x_1}^2 + {x_2}^2 - 3{x_1}{x_2} = 2{m^2} + \left| {m - 3} \right|\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 5{x_1}{x_2} = 2{m^2} + \left| {m - 3} \right|\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 5{x_1}{x_2} = 2{m^2} + \left| {m - 3} \right|\\
\to 4 - 5\left( {m - 1} \right) - 2{m^2} = \left| {m - 3} \right|\\
\to \left| {m - 3} \right| = - 2{m^2} - 5m + 9\\
\to \left[ \begin{array}{l}
m - 3 = - 2{m^2} - 5m + 9\left( {m \ge 3} \right)\\
- m + 3 = - 2{m^2} - 5m + 9\left( {m < 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2{m^2} + 6m - 12 = 0\\
2{m^2} + 4m - 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = \dfrac{{ - 3 + \sqrt {33} }}{2}\left( l \right)\\
m = \dfrac{{ - 3 - \sqrt {33} }}{2}\left( l \right)\\
m = 1\\
m = - 3
\end{array} \right.
\end{array}\)
