Đáp án:
\(m = \dfrac{{5 + \sqrt {17} }}{2}\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to {m^2} + 4m + 4 - {m^2} + 4 \ge 0\\
\to 4m + 8 \ge 0\\
\to m \ge - 2\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 4\\
{x_1}{x_2} = {m^2} - 4
\end{array} \right.\\
\dfrac{{{x_1}}}{{{x_2}}} - \dfrac{{{x_2}}}{{{x_1}}} = 8\\
\to \dfrac{{{x_1}^2 - {x_2}^2}}{{{x_1}{x_2}}} = 8\\
\to \dfrac{{\left( {{x_1} - {x_2}} \right)\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}} = 8\\
\to \left( {2m + 4} \right)\left( {{x_1} - {x_2}} \right) = 8\left( {{m^2} - 4} \right)\left( {DK:m \ne \pm 2} \right)\\
\to 2\left( {m + 2} \right)\left( {{x_1} - {x_2}} \right) = 8\left( {m - 2} \right)\left( {m + 2} \right)\\
\to \left( {m + 2} \right)\left( {{x_1} - {x_2} - 4\left( {m - 2} \right)} \right) = 0\\
\to \left[ \begin{array}{l}
m = - 2\left( l \right)\\
{x_1} - {x_2} = 4\left( {m - 2} \right)
\end{array} \right.\\
\to {x_1}^2 - 2{x_1}{x_2} + {x_2}^2 = 16\left( {{m^2} - 4m + 4} \right)\left( {DK:m > 2} \right)\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2} = 16{m^2} - 64m + 64\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 16{m^2} - 64m + 64\\
\to 4{m^2} + 16m + 16 - 4\left( {{m^2} - 4} \right) = 16{m^2} - 64m + 64\\
\to 16{m^2} - 80m + 32 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{{5 + \sqrt {17} }}{2}\\
m = \dfrac{{5 - \sqrt {17} }}{2}\left( l \right)
\end{array} \right.
\end{array}\)
