Đáp án:
\(\left[ \begin{array}{l}
m = - 4\\
m = 1
\end{array} \right.\)
Giải thích các bước giải:
2) Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta > 0\\
\to 4{m^2} + 8m + 4 - 4\left( { - 2m - 3} \right) > 0\\
\to 4{m^2} + 16m + 16 > 0\\
\to {\left( {2m + 4} \right)^2} > 0\\
\to m \ne - 2\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2m - 2\\
{x_1}{x_2} = - 2m - 3
\end{array} \right.\\
\left| {{x_1}} \right| + \left| {{x_2}} \right| = 6\\
\to {x_1}^2 + 2\left| {{x_1}{x_2}} \right| + {x_2}^2 = 36\\
\to \left[ \begin{array}{l}
{x_1}^2 + 2{x_1}{x_2} + {x_2}^2 = 36\\
{x_1}^2 - 2{x_1}{x_2} + {x_2}^2 = 36\left( {DK:1.\left( { - 2m - 3} \right) < 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{\left( {{x_1} + {x_2}} \right)^2} = 36\left( {DK:m \le - \dfrac{3}{2}} \right)\\
{x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2} = 36\left( {DK:m > - \dfrac{3}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{\left( { - 2m - 2} \right)^2} = 36\\
{\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 36
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 2m - 2 = 6\\
- 2m - 2 = - 6\\
4{m^2} + 8m + 4 - 4.\left( { - 2m - 3} \right) = 36
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - 4\\
m = 2\left( l \right)\\
4{m^2} + 16m - 20 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - 4\\
m = 1\\
m = - 5\left( l \right)
\end{array} \right.
\end{array}\)
