Đáp án:
m=4
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 1 - m + 3 \ge 0\\
\to 4 \ge m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 3
\end{array} \right.\\
\end{array}\)
\(\begin{array}{l}
\left[ \begin{array}{l}
x = 1 + \sqrt {4 - m} \\
x = 1 - \sqrt {4 - m}
\end{array} \right.\\
{x_1}^2 + {x_2}^2 + 2{x_2}^2 = 4{x_1}{x_2}\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} + 2{x_2}^2 = 6{x_1}{x_2}\\
\to {\left( {{x_1} + {x_2}} \right)^2} + 2{x_2}^2 = 6{x_1}{x_2}\\
\to \left[ \begin{array}{l}
4 + 2.{\left( {1 + \sqrt {4 - m} } \right)^2} = 6.\left( {m - 3} \right)\\
4 + 2.{\left( {1 - \sqrt {4 - m} } \right)^2} = 6.\left( {m - 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\left( {1 + 2\sqrt {4 - m} + 4 - m} \right) = 6m - 22\\
2\left( {1 - 2\sqrt {4 - m} + 4 - m} \right) = 6m - 22
\end{array} \right.\\
\to \left[ \begin{array}{l}
5 - m + 2\sqrt {4 - m} = 3m - 11\\
5 - m - 2\sqrt {4 - m} = 3m - 11
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt {4 - m} = 4m - 16\\
2\sqrt {4 - m} - 4m + 16
\end{array} \right.\\
\to \sqrt {4 - m} = 2m - 8\\
\to 4 - m = 4{m^2} - 32m + 64\left( {DK:4 \ge m \ge 4} \right)\\
\to 4{m^2} - 31m + 60 = 0\\
\to \left[ \begin{array}{l}
m = 4\\
m = \dfrac{{15}}{4}\left( l \right)
\end{array} \right.
\end{array}\)
