Đáp án: a. m=7 hoặc m=-3
b) thiếu đề
Giải thích các bước giải:
$\begin{array}{l}
{x^2} + \left( {m - 1} \right)x + m + 6 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta > 0\\
{x_1} + {x_2} = 1 - m\\
{x_1}{x_2} = m + 6
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\left( {m - 1} \right)^2} - 4m - 24 > 0\\
{x_1} + {x_2} = 1 - m\\
{x_1}{x_2} = m + 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 3 + 4\sqrt 2 \\
m < 3 - 4\sqrt 2
\end{array} \right.\\
{x_1} + {x_2} = 1 - m\\
{x_1}{x_2} = m + 6
\end{array} \right.\\
x_1^2 + x_2^2 = 10\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 10\\
\Rightarrow {\left( {1 - m} \right)^2} - 2\left( {m + 6} \right) = 10\\
\Rightarrow {m^2} - 4m - 21 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 7\\
m = - 3
\end{array} \right.\left( {tmdk} \right)\\
b)\\
2{x^2} - \left( {m + 3} \right)x + m - 1 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta > 0\\
{x_1} + {x_2} = \frac{{m + 3}}{2}\\
{x_1}{x_2} = \frac{{m - 1}}{2}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\left( {m + 3} \right)^2} - 8\left( {m - 1} \right) > 0\\
{x_1} + {x_2} = \frac{{m + 3}}{2}\\
{x_1}{x_2} = \frac{{m - 1}}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \in R\\
{x_1} + {x_2} = \frac{{m + 3}}{2}\\
{x_1}{x_2} = \frac{{m - 1}}{2}
\end{array} \right.\\
\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = ?
\end{array}$
