Đáp án:
\(\left[ \begin{array}{l}
m = \sqrt {2 + \sqrt 5 } \\
m = - \sqrt {2 + \sqrt 5 }
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\Leftrightarrow \Delta ' \ge 0\\
\to {m^2} - 4 \ge 0\\
\to \left[ \begin{array}{l}
m \ge 2\\
m \le - 2
\end{array} \right.\\
Có:{\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right)^2} + {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} = 3\\
\to \dfrac{{{x_1}^4 + {x_2}^4}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} = 3\\
\to \dfrac{{\left( {{x_1}^4 + 2{x_1}^2{x_2}^2 + {x_2}^4} \right) - 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} = 3\\
\to \dfrac{{{{\left( {{x_1}^2 + {x_2}^2} \right)}^2} - 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} = 3\\
\to \dfrac{{{{\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right]}^2} - 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} = 3\\
\to \dfrac{{{{\left( {4{m^2} - 2.4} \right)}^2} - 2.16}}{{16}} = 3\\
\to {\left( {4{m^2} - 2.4} \right)^2} = 80\\
\to \left[ \begin{array}{l}
4{m^2} - 2.4 = 4\sqrt 5 \\
4{m^2} - 2.4 = - 4\sqrt 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
{m^2} = 2 + \sqrt 5 \\
{m^2} = - 2 + \sqrt 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = \sqrt {2 + \sqrt 5 } \\
m = - \sqrt {2 + \sqrt 5 } \\
m = \sqrt { - 2 + \sqrt 5 } \left( l \right)\\
m = - \sqrt { - 2 + \sqrt 5 } \left( l \right)
\end{array} \right.
\end{array}\)
