Đáp án:
$\begin{array}{l}
a)m = - 1\\
\Rightarrow {x^2} + 2x - 3 = 0\\
\Rightarrow {x^2} + 2x + 1 = 4\\
\Rightarrow {\left( {x + 1} \right)^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\\
b){x_1} = - 1\\
\Rightarrow {\left( { - 1} \right)^2} + 2.\left( { - 1} \right) + m - 2 = 0\\
\Rightarrow m = 3\\
Do:{x_1} + {x_2} = - 2\\
\Rightarrow {x_2} = - 2 - \left( { - 1} \right) = - 1\\
c)2\,nghiệm\,trái\,dấu\\
\Rightarrow a.c < 0\\
\Rightarrow 1.\left( {m - 2} \right) < 0\\
\Rightarrow m < 2\\
d)\Delta ' \ge 0\\
\Rightarrow 1 - m + 2 \ge 0\\
\Rightarrow m \le 3\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
{x_1}{x_2} = m - 2
\end{array} \right.\\
M = {\left( {{x_1} - {x_2}} \right)^2} + 10{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} + 10{x_1}{x_2}\\
= {\left( { - 2} \right)^2} + 6.\left( {m - 2} \right)\\
= 6m - 8\\
Do:m \le 3\\
\Rightarrow 6m \le 18\\
\Rightarrow 6m - 8 \le 10\\
\Rightarrow M \le 10\\
\Rightarrow {\mathop{\rm maxM}\nolimits} = 10 \Leftrightarrow m = 3
\end{array}$
