Đáp án:
$\begin{array}{l}
1){x^2} - \left( {3m + 1} \right)x + 2{m^2} + m = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {\left( {3m + 1} \right)^2} - 4\left( {2{m^2} + m} \right) > 0\\
\Rightarrow 9{m^2} + 6m + 1 - 8{m^2} - 4m > 0\\
\Rightarrow {m^2} + 2m + 1 > 0\\
\Rightarrow m \ne - 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3m + 1\\
{x_1}{x_2} = 2{m^2} + m
\end{array} \right.\\
\left| {{x_1} - {x_2}} \right| = 1\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 1\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 1\\
\Rightarrow {\left( {3m + 1} \right)^2} - 4\left( {2{m^2} + m} \right) = 1\\
\Rightarrow {m^2} + 2m + 1 = 1\\
\Rightarrow m\left( {m + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 0\left( {tm} \right)\\
m = - 2\left( {tm} \right)
\end{array} \right.\\
2){x^2} - \left( {m + 2} \right).x + m - 1 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {\left( {m + 2} \right)^2} - 4m + 4 > 0\\
\Rightarrow {m^2} + 4m + 4 - 4m + 4 > 0\\
\Rightarrow {m^2} + 8 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 2\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
A = x_1^2 + x_2^2 - 4{x_1}{x_2}\\
= \left( {{x_1} + {x_2}} \right) - 6{x_1}{x_2}\\
= {\left( {m + 2} \right)^2} - 6.\left( {m - 1} \right)\\
= {m^2} + 4m + 4 - 6m + 6\\
= {m^2} - 2m + 1 + 9\\
= {\left( {m - 1} \right)^2} + 9 \ge 9\\
\Rightarrow GTNN:A = 9\\
Khi:m = 1
\end{array}$
Vậy m=1
