Đáp án:
\(m = \dfrac{3}{2}\)
Giải thích các bước giải:
Để phương trình có nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to 4{m^2} - 4m + 1 - 4.2\left( {m - 1} \right) \ge 0\\
\to 4{m^2} - 12m + 9 \ge 0\\
\to {\left( {2m - 3} \right)^2} \ge 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{ - 2m + 1}}{2}\\
{x_1}{x_2} = \dfrac{{m - 1}}{2}
\end{array} \right.\\
\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} = - 4\\
\to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = - 4\\
\to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = \dfrac{{ - 4{x_1}{x_2}}}{{{x_1}{x_2}}}\\
\to \dfrac{{{x_1} + {x_2} + 4{x_1}{x_2}}}{{{x_1}{x_2}}} = 0\\
\to \left\{ \begin{array}{l}
{x_1} + {x_2} + 4{x_1}{x_2} = 0\\
{x_1}{x_2} \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{ - 2m + 1}}{2} + 4.\dfrac{{m - 1}}{2} = 0\\
\dfrac{{m - 1}}{2} \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
- 2m + 1 + 4m - 4 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
2m = 3
\end{array} \right.\\
\to m = \dfrac{3}{2}
\end{array}\)
