Đáp án:
m=1
Giải thích các bước giải:
\(\begin{array}{l}
{x^2} - x - 3m + 1 = 0\\
Xét:\Delta = 1 - 4\left( { - 3m + 1} \right) \ge 0\\
\to 1 + 12m - 4 \ge 0\\
\to 12m - 3 \ge 0\\
\to m \ge \dfrac{1}{4}\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {12m - 3} }}{2}\\
x = \dfrac{{1 - \sqrt {12m - 3} }}{2}
\end{array} \right.\\
Có:{x_1} - 2{x_2} = 4\\
\to {x_1} + {x_2} - 3{x_2} = 4\\
\to \left[ \begin{array}{l}
1 - 3\left( {\dfrac{{1 + \sqrt {12m - 3} }}{2}} \right) = 4\\
1 - 3\left( {\dfrac{{1 - \sqrt {12m - 3} }}{2}} \right) = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
2 - 3 - 3\sqrt {12m - 3} = 8\\
2 - 3 + 3\sqrt {12m - 3} = 8
\end{array} \right.\\
\to \left[ \begin{array}{l}
3\sqrt {12m - 3} = - 9\left( l \right)\\
3\sqrt {12m - 3} = 9
\end{array} \right.\\
\to \sqrt {12m - 3} = 3\\
\to 12m - 3 = 9\\
\to 12m = 12\\
\to m = 1\left( {TM} \right)
\end{array}\)
