Đáp án:
\(\left[ \begin{array}{l}
m = - 2\\
m = 6 + 4\sqrt 2 \\
m = 6 - 4\sqrt 2
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to {m^2} - 4m + 4 - {m^2} + 2m - 4 > 0\\
\to - 2m > 0\\
\to m < 0\\
Có:\dfrac{2}{{{x_1}^2 + {x_2}^2}} - \dfrac{1}{{{x_1}{x_2}}} = \dfrac{1}{{15m}}\\
\to \dfrac{2}{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}} - \dfrac{1}{{{x_1}{x_2}}} = \dfrac{1}{{15m}}\\
\to \dfrac{2}{{{{\left( {4 - 2m} \right)}^2} - 2\left( {{m^2} - 2m + 4} \right)}} - \dfrac{1}{{{m^2} - 2m + 4}} = \dfrac{1}{{15m}}\\
\to \dfrac{2}{{16 - 16m + 4{m^2} - 2{m^2} + 4m - 8}} - \dfrac{1}{{{m^2} - 2m + 4}} = \dfrac{1}{{15m}}\\
\to \dfrac{2}{{2{m^2} - 12m + 8}} - \dfrac{1}{{{m^2} - 2m + 4}} = \dfrac{1}{{15m}}\\
\to \dfrac{1}{{{m^2} - 6m + 4}} - \dfrac{1}{{{m^2} - 2m + 4}} = \dfrac{1}{{15m}}\\
\to \dfrac{{{m^2} - 2m + 4 - {m^2} + 6m - 4}}{{\left( {{m^2} - 6m + 4} \right)\left( {{m^2} - 2m + 4} \right)}} = \dfrac{1}{{15m}}\\
\to \dfrac{{4m}}{{\left( {{m^2} - 6m + 4} \right)\left( {{m^2} - 2m + 4} \right)}} = \dfrac{1}{{15m}}\\
\to {m^4} - 2{m^3} + 4{m^2} - 6{m^3} + 12{m^2} - 24m + 4{m^2} - 8m + 16 = 60{m^2}\\
\to {m^4} - 8{m^3} - 40{m^2} - 32m + 16 = 0\\
\to {m^4} + 2{m^3} - 10{m^3} - 20{m^2} - 20{m^2} - 40m + 8m + 16 = 0\\
\to {m^3}\left( {m + 2} \right) - 10{m^2}\left( {m + 2} \right) - 20m\left( {m + 2} \right) + 8\left( {m + 2} \right) = 0\\
\to \left[ \begin{array}{l}
m + 2 = 0\\
{m^3} - 10{m^2} - 20m + 8 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - 2\\
\left( {m + 2} \right)\left( {{m^2} - 12m + 4} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - 2\\
m = 6 + 4\sqrt 2 \\
m = 6 - 4\sqrt 2
\end{array} \right.
\end{array}\)
