Đáp án:
$\begin{array}{l}
a)m = - 1\\
\Rightarrow {x^2} + 4x + 1 = 0\\
\Rightarrow {x^2} + 4x + 4 - 3 = 0\\
\Rightarrow {\left( {x + 2} \right)^2} = 3\\
\Rightarrow \left[ \begin{array}{l}
x = - 2 + \sqrt 3 \\
x = - 2 - \sqrt 3
\end{array} \right.\\
b)\Delta ' > 0\\
\Rightarrow {\left( {m - 1} \right)^2} - m - 2 > 0\\
\Rightarrow {m^2} - 2m + 1 - m - 2 > 0\\
\Rightarrow {m^2} - 3m - 1 > 0\\
\Rightarrow {m^2} - 2.m.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{{13}}{4} > 0\\
\Rightarrow {\left( {m - \dfrac{3}{2}} \right)^2} > \dfrac{{13}}{4}\\
\Rightarrow \left[ \begin{array}{l}
m > \dfrac{{3 + \sqrt {13} }}{2}\\
m < \dfrac{{3 - \sqrt {13} }}{2}
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = m + 2
\end{array} \right.\\
\left| {{x_1} - {x_2}} \right| = 2\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 4\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 4\\
\Rightarrow 4{\left( {m - 1} \right)^2} - 4.\left( {m + 2} \right) = 4\\
\Rightarrow {m^2} - 2m + 1 - m - 2 = 1\\
\Rightarrow {m^2} - 3m - 2 = 0\\
\Rightarrow m = \dfrac{{3 \pm \sqrt {17} }}{2}\left( {tmdk} \right)\\
c)\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = m + 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 2\\
2{x_1}{x_2} = 2m + 4
\end{array} \right.\\
\Rightarrow 2{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) = 6
\end{array}$
