Đáp án:
\(\left[ \begin{array}{l}
m = \dfrac{{7 + \sqrt {57} }}{4}\\
m = \dfrac{{7 - \sqrt {57} }}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:m = - 1\\
Pt \to {x^2} + 4x = 0\\
\to x\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 4
\end{array} \right.
\end{array}\)
b. Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to {m^2} - 2m + 1 - m - 1 \ge 0\\
\to {m^2} - 3m \ge 0\\
\to m\left( {m - 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
m \le 0\\
m \ge 3
\end{array} \right.\\
Có:\dfrac{{{x_1}}}{{{x_2}}} + \dfrac{{{x_2}}}{{{x_1}}} = 4\\
\to \dfrac{{{x_1}^2 + {x_2}^2}}{{{x_1}{x_2}}} = 4\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2} = 4{x_1}{x_2}\\
\to {\left( {{x_1} + {x_2}} \right)^2} = 6{x_1}{x_2}\\
\to {\left( {2m - 2} \right)^2} - 6\left( {m + 1} \right) = 0\\
\to 4{m^2} - 8m + 4 - 6m - 6 = 0\\
\to 4{m^2} - 14m - 2 = 0\\
\to 2{m^2} - 7m - 1 = 0
\end{array}\)
Xét:
\(\begin{array}{l}
Δ= 49 - 4.2.\left( { - 1} \right) = 57\\
\to \left[ \begin{array}{l}
m = \dfrac{{7 + \sqrt {57} }}{4}\\
m = \dfrac{{7 - \sqrt {57} }}{4}
\end{array} \right.\left( {TM} \right)
\end{array}\)
