Đáp án:
\(m = \dfrac{{15 + \sqrt {33} }}{8}\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 1 - m + 3 \ge 0\\
\to 4 \ge m\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt {4 - m} \\
x = 1 - \sqrt {4 - m}
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 3
\end{array} \right.\\
{x_1}^2 + 3{x_2}^2 = 4{x_1}{x_2}\\
\to {x_1}^2 + {x_2}^2 + 2{x_2}^2 = 4{x_1}{x_2}\\
\to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) + 2{x_2}^2 = 6{x_1}{x_2}\\
\to {\left( {{x_1} + {x_2}} \right)^2} + 2{x_2}^2 = 6{x_1}{x_2}\\
\to \left[ \begin{array}{l}
4 + 2.{\left( {1 + \sqrt {4 - m} } \right)^2} = 6\left( {m - 3} \right)\\
4 + 2.{\left( {1 - \sqrt {4 - m} } \right)^2} = 6\left( {m - 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
4 + 2\left( {1 + 2\sqrt {4 - m} + 4 - m} \right) = 6m - 18\\
4 + 2\left( {1 - 2\sqrt {4 - m} + 4 - m} \right) = 6m - 18
\end{array} \right.\\
\to \left[ \begin{array}{l}
14 - 2m + 4\sqrt {4 - m} = 6m - 18\\
14 - 2m - 4\sqrt {4 - m} = 6m - 18
\end{array} \right.\\
\to \left[ \begin{array}{l}
4\sqrt {4 - m} = 8m - 32\\
4\sqrt {4 - m} = - 8m + 32
\end{array} \right.\\
\to \sqrt {4 - m} = 2m - 4\\
\to 4 - m = 4{m^2} - 16m + 16\left( {DK:4 \ge m \ge 2} \right)\\
\to 4{m^2} - 15m + 12 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{{15 + \sqrt {33} }}{8}\left( {TM} \right)\\
m = \dfrac{{15 - \sqrt {33} }}{8}\left( l \right)
\end{array} \right.
\end{array}\)
