Đáp án:
\(m = -\dfrac14\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to {m^2} + m + 4 > 0\\
\to {m^2} + 2m.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{15}}{4} > 0\\
\to {\left( {m + \dfrac{1}{2}} \right)^2} + \dfrac{{15}}{4} > 0\left( {ld} \right)\forall m\\
Có:\dfrac{1}{{{x_1}^2 + {x_2}^2}}\\
= \dfrac{1}{{{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}}}\\
= \dfrac{1}{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}}\\
= \dfrac{1}{{4{m^2} - 2\left( { - m - 4} \right)}}\\
= \dfrac{1}{{4{m^2} + 2m + 8}}\\
= \dfrac{1}{{\left( {4{m^2} + 2.2m.\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{{31}}{4}}}\\
= \dfrac{1}{{{{\left( {2m + \dfrac{1}{2}} \right)}^2} + \dfrac{{31}}{4}}}\\
Do:{\left( {2m + \dfrac{1}{2}} \right)^2} \ge 0\forall m\\
\to {\left( {2m + \dfrac{1}{2}} \right)^2} + \dfrac{{31}}{4} \ge \dfrac{{31}}{4}\\
\to \dfrac{1}{{{{\left( {2m + \dfrac{1}{2}} \right)}^2} + \dfrac{{31}}{4}}} \le \dfrac{4}{{31}}\\
\to Max = \dfrac{4}{{31}}\\
\Leftrightarrow 2m + \dfrac{1}{2} = 0\\
\to m = - \dfrac{1}{4}
\end{array}\)
