Đáp án:
\(\left[ \begin{array}{l}
m > \sqrt 5 \\
m < - \sqrt 5
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to {m^2} - 4 > 0\\
\to \left( {m - 2} \right)\left( {m + 2} \right) > 0\\
\to \left[ \begin{array}{l}
m > 2\\
m < - 2
\end{array} \right.\\
Có:\dfrac{{{x_1}^2}}{{{x_2}^2}} + \dfrac{{{x_2}^2}}{{{x_1}^2}} > 7\\
\to \dfrac{{{x_1}^4 + {x_2}^4}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{\left( {{x_1}^4 + {x_2}^4 + 2{x_1}^2{x_1}^2} \right) - 2{x_1}^2{x_2}^2}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{{\left( {{x_1}^2 + {x_2}^2} \right)}^2} - 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{{\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right]}^2} - 2{{\left( {{x_1}{x_2}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} > 7\\
\to \dfrac{{{{\left[ {{m^2} - 2} \right]}^2} - 2.1}}{1} > 7\\
\to {m^4} - 4{m^2} + 4 - 2 > 7\\
\to {m^4} - 4{m^2} - 5 > 0\\
\to \left( {{m^2} - 5} \right)\left( {{m^2} + 1} \right) > 0\\
\to {m^2} - 5 > 0\left( {do:{m^2} + 1 > 0\forall m \in R} \right)\\
\to \left[ \begin{array}{l}
m > \sqrt 5 \\
m < - \sqrt 5
\end{array} \right.\\
KL:\left[ \begin{array}{l}
m > \sqrt 5 \\
m < - \sqrt 5
\end{array} \right.
\end{array}\)
