Đáp án:
\(m \in \emptyset \)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta > 0\\
\to {m^2} - 2m + 1 - 4m > 0\\
\to {m^2} - 6m + 1 > 0\\
\to {m^2} - 6m + 9 > 8\\
\to {\left( {m - 3} \right)^2} > 8\\
\to \left[ \begin{array}{l}
m - 3 > 2\sqrt 2 \\
m - 3 < - 2\sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
m > 3 + 2\sqrt 2 \\
m < - 3 - 2\sqrt 2
\end{array} \right.\\
\to Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 1\\
{x_1}{x_2} = - m
\end{array} \right.\\
\left| {{x_1}} \right| = \left| {{x_2}} \right| + 1\\
\to \left| {{x_1}} \right| - \left| {{x_2}} \right| = 1\\
\to {x_1}^2 - 2\left| {{x_1}{x_2}} \right| + {x_2}^2 = 1\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 2{x_1}{x_2} - 2\left| {{x_1}{x_2}} \right| = 1\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} - 2\left| {{x_1}{x_2}} \right| = 1\\
\to \left[ \begin{array}{l}
{\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 1\left( {DK:m \le 0} \right)\\
{\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + 2{x_1}{x_2} = 1\left( {DK:1.\left( { - m} \right) < 0 \to m > 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{m^2} - 2m + 1 - 4\left( { - m} \right) = 1\\
{m^2} - 2m + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
{m^2} + 2m = 0\\
{m^2} - 2m = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m\left( {m + 2} \right) = 0\\
m\left( {m - 2} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 0\left( {KTM} \right)\\
m = - 2\left( {KTM} \right)\\
m = 2\left( {KTM} \right)
\end{array} \right.
\end{array}\)
\( \to m \in \emptyset \)
