Đáp án:
m=-7
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to 1 - 4\left( {m + 1} \right) > 0\\
\to 1 - 4m - 4 > 0\\
\to - \dfrac{3}{4} > m\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt { - 3 - 4m} }}{2}\\
x = \dfrac{{1 - \sqrt { - 3 - 4m} }}{2}
\end{array} \right.\\
{x_1}^2 + {x_1}{x_2} + 3{x_2} = 7\\
\to \left[ \begin{array}{l}
{\left( {\dfrac{{1 + \sqrt { - 3 - 4m} }}{2}} \right)^2} + m + 1 + 3\left( {\dfrac{{1 - \sqrt { - 3 - 4m} }}{2}} \right) = 7\\
{\left( {\dfrac{{1 - \sqrt { - 3 - 4m} }}{2}} \right)^2} + m + 1 + 3\left( {\dfrac{{1 + \sqrt { - 3 - 4m} }}{2}} \right) = 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{1 + 2\sqrt { - 3 - 4m} - 3 - 4m}}{4} + m + 1 + \dfrac{{3 - 3\sqrt { - 3 - 4m} }}{2} = 7\\
\dfrac{{1 - 2\sqrt { - 3 - 4m} - 3 - 4m}}{4} + m + 1 + \dfrac{{3 + 3\sqrt { - 3 - 4m} }}{2} = 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{1 + 2\sqrt { - 3 - 4m} - 3 - 4m + 4m + 4 + 6 - 6\sqrt { - 3 - 4m} - 28}}{4} = 0\\
\dfrac{{1 - 2\sqrt { - 3 - 4m} - 3 - 4m + 4m + 4 + 6 + 6\sqrt { - 3 - 4m} - 28}}{4} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 4\sqrt { - 3 - 4m} - 20 = 0\left( l \right)\\
4\sqrt { - 3 - 4m} - 20 - 0
\end{array} \right.\\
\to \sqrt { - 3 - 4m} = 5\\
\to - 3 - 4m = 25\\
\to m = - 7
\end{array}\)
