Đáp án:
b. \(m = \pm \frac{3}{{\sqrt 2 }}\)
Giải thích các bước giải:
a. Để phương trình luôn có 2 nghiệm trái dấu
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {m + 1} \right)^2} + 4 > 0\left( {ld} \right)\forall m \in R\\
- 4.1 < 0\left( {ld} \right)
\end{array} \right.\\
\to dpcm
\end{array}\)
b. Xét:
\(\begin{array}{l}
\frac{{{x_1}^2.{x_2}^2 - 2{x_1}^2.{x_2} - 4{x_1}^2 - 2{x_1}{x_2}^2 + 2{x_1}{x_2} + 8{x_1} - 4{x_2}^2 + 8{x_2} + 16}}{{{x_1}{x_2}}} = 16\\
\to {\left( {{x_1}{x_2}} \right)^2} - 2{x_1}{x_2}\left( {{x_1} + {x_2}} \right) - 4\left( {{x_1}^2 + {x_2}^2} \right) + 2{x_1}{x_2} + 8\left( {{x_1} + {x_2}} \right) + 16 = 16{x_1}{x_2}\\
\to {\left( {{x_1}{x_2}} \right)^2} - 2{x_1}{x_2}\left( {{x_1} + {x_2}} \right) - 4\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) + 8\left( {{x_1} + {x_2}} \right) - 14{x_1}{x_2} + 16 = 0\\
\to {\left( {{x_1}{x_2}} \right)^2} - 2{x_1}{x_2}\left( {{x_1} + {x_2}} \right) - 4{\left( {{x_1} + {x_2}} \right)^2} + 8{x_1}{x_2} + 8\left( {{x_1} + {x_2}} \right) - 14{x_1}{x_2} + 16 = 0\\
\to {\left( {{x_1}{x_2}} \right)^2} - 2{x_1}{x_2}\left( {{x_1} + {x_2}} \right) - 4{\left( {{x_1} + {x_2}} \right)^2} - 6{x_1}{x_2} + 8\left( {{x_1} + {x_2}} \right) + 16 = 0\\
\to {\left( { - 4} \right)^2} - 2.\left( { - 4} \right).\left( {2m + 2} \right) - 4{\left( {2m + 2} \right)^2} - 6.\left( { - 4} \right) + 8\left( {2m + 2} \right) + 16 = 0\\
\to 16 + 8\left( {2m + 2} \right) - 4\left( {4{m^2} + 8m + 4} \right) + 24 + 16m + 16 + 16 = 0\\
\to 16 + 16m + 16 - 16{m^2} - 32m - 16 + 24 + 16m + 16 + 16 = 0\\
\to - 16{m^2} + 72 = 0\\
\to \frac{9}{2} - {m^2} = 0\\
\to {m^2} = \frac{9}{2}\\
\to m = \pm \frac{3}{{\sqrt 2 }}
\end{array}\)
