Đáp án:
\(\left[ \begin{array}{l}
m = - 5\\
m = - 1
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta > 0\\
\to 1 - 2m + {m^2} - 4\left( { - 2m - 4} \right) > 0\\
\to {m^2} + 6m + 17 > 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 1\\
{x_1}{x_2} = - 2m - 4
\end{array} \right.\\
\dfrac{1}{{{{\left( {{x_1} + 2} \right)}^2}}} + \dfrac{1}{{{{\left( {{x_2} + 2} \right)}^2}}} = 2\\
\to \dfrac{{{x_1}^2 + 4{x_1} + 4 + {x_2}^2 + 4{x_2} + 4}}{{{{\left[ {\left( {{x_1} + 2} \right)\left( {{x_2} + 2} \right)} \right]}^2}}} = 2\\
\to \dfrac{{{x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 2{x_1}{x_2} + 4\left( {{x_1} + {x_2}} \right) + 8}}{{{{\left( {{x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right) + 4} \right)}^2}}} = 2\\
\to \dfrac{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2} + 4\left( {{x_1} + {x_2}} \right) + 8}}{{{{\left( {{x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right) + 4} \right)}^2}}} = 2\\
\to \dfrac{{{m^2} - 2m + 1 - 2\left( { - 2m - 4} \right) + 4\left( {m - 1} \right) + 8}}{{{{\left( { - 2m - 4 + 2\left( {m - 1} \right) + 4} \right)}^2}}} = 2\\
\to \dfrac{{{m^2} + 6m + 13}}{4} = 2\\
\to {m^2} + 6m + 13 = 8\\
\to {m^2} + 6m + 5 = 0\\
\to \left( {m + 1} \right)\left( {m + 5} \right) = 0\\
\to \left[ \begin{array}{l}
m = - 5\\
m = - 1
\end{array} \right.
\end{array}\)
