Đáp án:
$\begin{array}{l}
m + 1 \ne 0 \Rightarrow m \ne - 1\\
\left( {m + 1} \right){x^2} + \left( {4 - 2m} \right).x + 4 - 2m = 0\\
\Rightarrow \left( {m + 1} \right){x^2} + 2.\left( {2 - m} \right).x + 4 - 2m = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow {\left( {2 - m} \right)^2} - \left( {m + 1} \right)\left( {4 - 2m} \right) > 0\\
\Rightarrow \left( {2 - m} \right)\left( {2 - m - 2m - 2} \right) > 0\\
\Rightarrow \left( {2 - m} \right).\left( { - 3m} \right) > 0\\
\Rightarrow 3m\left( {m - 2} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
m > 2\\
m < 0
\end{array} \right.
\end{array}$
Khi ngăn cách giữa x1 và x2 bằng 3 thì:
$\begin{array}{l}
\left| {{x_1} - {x_2}} \right| = 3\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 9\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 9\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2m - 4}}{{m + 1}}\\
{x_1}{x_2} = \dfrac{{4 - 2m}}{{m + 1}}
\end{array} \right.\\
\Rightarrow {\left( {\dfrac{{2m - 4}}{{m + 1}}} \right)^2} - 4.\dfrac{{4 - 2m}}{{m + 1}} = 9\\
\Rightarrow \dfrac{{4{m^2} - 16m + 16 + \left( {8m - 16} \right).\left( {m + 1} \right)}}{{{{\left( {m + 1} \right)}^2}}} = 9\\
\Rightarrow 4{m^2} - 16m + 16 + 8{m^2} + 8m - 16m - 16\\
= 9{m^2} + 18m + 9\\
\Rightarrow 3{m^2} - 42m - 9 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 7 + 2\sqrt {13} \left( {tm} \right)\\
m = 7 - 2\sqrt {13} \left( {tm} \right)
\end{array} \right.\\
Vậy\,m = 7 \pm 2\sqrt {13}
\end{array}$
