Đáp án: $m = 1;m = \dfrac{{ - 3}}{5}$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} - \left( {m - 1} \right)x - 1 - {m^2} = 0\\
\Delta > 0\\
\Rightarrow {\left( {m - 1} \right)^2} - 4.\left( { - 1 - {m^2}} \right) > 0\\
\Rightarrow {\left( {m - 1} \right)^2} + 4{m^2} + 4 > 0\left( {luôn\,đúng} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 1\\
{x_1}{x_2} = - 1 - {m^2}
\end{array} \right.\\
\left| {{x_1}} \right| + \left| {{x_2}} \right| = 2\sqrt 2 \\
\Rightarrow x_1^2 + x_2^2 + 2.\left| {{x_1}{x_2}} \right| = 8\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + 2\left| {{x_1}{x_2}} \right| = 8\\
\Rightarrow {\left( {m - 1} \right)^2} - 2.\left( { - 1 - {m^2}} \right) + 2.\left| { - 1 - {m^2}} \right| = 8\\
\Rightarrow {m^2} - 2m + 1 + 2 + 2{m^2} + 2.\left( {{m^2} + 1} \right) = 8\\
\Rightarrow 5{m^2} - 2m - 3 = 0\\
\Rightarrow 5{m^2} - 5m + 3m - 3 = 0\\
\Rightarrow \left( {m - 1} \right)\left( {5m + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 1\\
m = - \dfrac{3}{5}
\end{array} \right.
\end{array}$
Vậy $m = 1;m = \dfrac{{ - 3}}{5}$
