Đáp án:
\(\left[ \begin{array}{l}
m = \dfrac{{1 + \sqrt {145} }}{{18}}\\
m = \dfrac{{1 - \sqrt {145} }}{{18}}
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình \({x^2} - 6mx + m = 0\) có nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 9{m^2} - m \ge 0\\
\to m\left( {9m - 1} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \ge 0\\
9m - 1 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m \le 0\\
9m - 1 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
m \ge \dfrac{1}{9}\\
m \le 0
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 6m\\
{x_1}{x_2} = m
\end{array} \right.\\
Có:{x_1} - {x_2} = 4\\
\to {x_1}^2 - 2{x_1}{x_2} + {x_2}^2 = 16\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2} = 16\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 16\\
\to 36{m^2} - 4m = 16\\
\to 9{m^2} - m - 4 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{{1 + \sqrt {145} }}{{18}}\\
m = \dfrac{{1 - \sqrt {145} }}{{18}}
\end{array} \right.
\end{array}\)
