Đáp án:
c) m=3
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 1\\
Pt \to - 6x - 4 = 0\\
\to x = - \dfrac{2}{3}\\
b)Thay:m = 3\\
Pt \to 2{x^2} - 10x - 2 = 0\\
\Delta ' = 25 + 4 = 29\\
\to \left[ \begin{array}{l}
x = \dfrac{{5 + \sqrt {29} }}{2}\\
x = \dfrac{{5 - \sqrt {29} }}{2}
\end{array} \right.
\end{array}\)
c) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
{m^2} + 4m + 4 - \left( {m - 1} \right)\left( {m - 5} \right) \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
{m^2} + 4m + 4 - {m^2} + 6m - 5 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
10m - 1 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
m \ge \dfrac{1}{{10}}
\end{array} \right.\\
Có:\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} - {x_1}{x_2} + 4 = 0\\
\to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} - {x_1}{x_2} + 4 = 0\\
\to {x_1} + {x_2} - {\left( {{x_1}{x_2}} \right)^2} + 4{x_1}{x_2} = 0\\
\to \dfrac{{2m + 4}}{{m - 1}} - \dfrac{{{{\left( {m - 5} \right)}^2}}}{{{{\left( {m - 1} \right)}^2}}} + 4.\dfrac{{m - 5}}{{m - 1}} = 0\\
\to \dfrac{{6m - 16}}{{m + 1}} = \dfrac{{{{\left( {m - 5} \right)}^2}}}{{{{\left( {m - 1} \right)}^2}}}\\
\to \left( {6m - 16} \right)\left( {m - 1} \right) = {m^2} - 10m + 25\\
\to 6{m^2} - 22m + 16 = {m^2} - 10m + 25\\
\to 5{m^2} - 12m - 9 = 0\\
\to \left[ \begin{array}{l}
m = 3\\
m = - \dfrac{3}{5}\left( l \right)
\end{array} \right.
\end{array}\)
