Đáp án:

d. \({x_1} + {x_2} - 3{x_1}{x_2} =  - 5\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left\{ \begin{array}{l}
m \ne 0\\
{m^2} + 6m + 9 - 4m\left( {2m + 1} \right) = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ne 0\\
 - 7{m^2} + 2m + 9 = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ne 0\\
\left[ \begin{array}{l}
m = \frac{9}{7}\\
m =  - 1
\end{array} \right.
\end{array} \right.\\
b.\left\{ \begin{array}{l}
m \ne 0\\
 - 7{m^2} + 2m + 9 > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ne 0\\
\left( {m + 1} \right)\left( {7m - 9} \right) > 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m + 1 > 0\\
7m - 9 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 1 < 0\\
7m - 9 > 0
\end{array} \right.\left( l \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - 1\\
m < \frac{9}{7}
\end{array} \right.\\
KL: - 1 < m < \frac{9}{7};m \ne 0\\
c.{x_1} - {x_2} = 2\\
 \to {x_1}^2 + {x_2}^2 - 2{x_1}{x_2} = 4\\
 \to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) - 2{x_1}{x_2} - 2{x_1}{x_2} = 4\\
 \to {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 4\\
 \to \frac{{{m^2} + 6m + 9}}{{{m^2}}} - 4.\frac{{2m + 1}}{m} = 4\\
 \to {m^2} + 6m + 9 - 8{m^2} - 4m - 4{m^2} = 0\\
 \to  - 11{m^2} + 2m + 9 = 0\\
 \to \left[ \begin{array}{l}
m = 1\\
m =  - \frac{9}{{11}}
\end{array} \right.\\
d.Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{m + 3}}{m} = 1 + \frac{3}{m}\\
{x_1}{x_2} = \frac{{2m + 1}}{m} = 2 + \frac{1}{m}
\end{array} \right.\\
 \to {x_1} + {x_2} - 3{x_1}{x_2} = 1 + \frac{3}{m} - 3\left( {2 + \frac{1}{m}} \right)\\
 = 1 - 6 =  - 5\\
KL:{x_1} + {x_2} - 3{x_1}{x_2} =  - 5
\end{array}\)