Đáp án:
$\begin{array}{l}
1){x^2} - 2\left( {m + 1} \right).x + 4m = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 4m > 0\\
\Rightarrow {\left( {m - 1} \right)^2} > 0\\
\Rightarrow m \ne 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = 4m
\end{array} \right.\\
Do:{x_1} = - 3{x_2}\\
\Rightarrow - 3{x_2} + {x_2} = 2\left( {m + 1} \right)\\
\Rightarrow - 2{x_2} = 2\left( {m + 1} \right)\\
\Rightarrow {x_2} = - \left( {m + 1} \right)\\
\Rightarrow {x_1} = - 3{x_2} = 3\left( {m + 1} \right)\\
\Rightarrow {x_1}{x_2} = - 3x_2^2 = - 3{\left( {m + 1} \right)^2}\\
\Rightarrow - 3{\left( {m + 1} \right)^2} = 4m\\
\Rightarrow - 3{m^2} - 6m - 3 - 4m = 0\\
\Rightarrow 3{m^2} + 10m + 3 = 0\\
\Rightarrow \left( {3m + 1} \right)\left( {m + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = - \frac{1}{3}\left( {tm} \right)\\
m = - 3\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = - \frac{1}{3};m = - 3\\
C2)\\
\Delta ' > 0\\
\Rightarrow {\left( { - 3} \right)^2} - m - 3 > 0\\
\Rightarrow 9 - m - 3 > 0\\
\Rightarrow m < 6\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1}{x_2} = m + 3
\end{array} \right.\\
Do:{x_2} = x_1^2\\
\Rightarrow x_1^2 + {x_1} = 6\\
\Rightarrow x_1^2 + {x_1} - 6 = 0\\
\Rightarrow \left( {{x_1} - 2} \right)\left( {{x_1} + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x_1} = 2 \Rightarrow {x_2} = 4\\
{x_1} = - 3 \Rightarrow {x_2} = 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m + 3 = 2.4 = 8 \Rightarrow m = 5\\
m + 3 = \left( { - 3} \right).9 = - 27 \Rightarrow m = - 30
\end{array} \right.\left( {tm} \right)\\
Vậy\,m = - 30;m = 5
\end{array}$
