Đáp án:
b) \(m = - \dfrac{1}{5}\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\left\{ \begin{array}{l}
m \ne - 1\\
1 - m - 1 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
m \le 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt m }}{{m + 1}}\\
x = \dfrac{{1 - \sqrt m }}{{m + 1}}
\end{array} \right.\\
a)Có:{x_1} - 3{x_2} = 1\\
\to {x_1} + {x_2} - 4{x_2} = 1\\
\to \left[ \begin{array}{l}
\dfrac{2}{{m + 1}} - 4.\left( {\dfrac{{1 + \sqrt m }}{{m + 1}}} \right) = 1\\
\dfrac{2}{{m + 1}} - 4.\left( {\dfrac{{1 - \sqrt m }}{{m + 1}}} \right) = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{ - 2 - 4\sqrt m }}{{m + 1}} = 1\\
\dfrac{{ - 2 + 4\sqrt m }}{{m + 1}} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 2 - 4\sqrt m = m + 1\\
- 2 + 4\sqrt m = m + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
4\sqrt m = - 3 - m\\
4\sqrt m = m + 3
\end{array} \right.\\
\to 16m = {m^2} + 6m + 9\left( {DK:m \ge 0} \right)\\
\to {m^2} - 10m + 9 = 0\\
\to \left[ \begin{array}{l}
m = 9\\
m = 1
\end{array} \right.\left( {KTM} \right)\\
b)Có:2{x_1} + 2{x_2} = 5\\
\to 2\left( {{x_1} + {x_2}} \right) = 5\\
\to 2\left( {\dfrac{2}{{m + 1}}} \right) = 5\\
\to 4 = 5m + 5\\
\to 5m = - 1\\
\to m = - \dfrac{1}{5}
\end{array}\)
