Đáp án:
$\begin{array}{l}
{x^2} - \left( {m + 1} \right)x + m - 4 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 4\left( {m - 4} \right) > 0\\
\Rightarrow {m^2} + 2m + 1 - 4m + 16 > 0\\
\Rightarrow {m^2} - 2m + 1 + 16 > 0\\
\Rightarrow {\left( {m - 1} \right)^2} + 16 > 0\left( {luon\,dung} \right)\\
TheoViet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 1\\
{x_1}{x_2} = m - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x_1^2 - \left( {m + 1} \right){x_1} + m - 4 = 0\\
x_2^2 - \left( {m + 1} \right){x_2} + m - 4 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x_1^2 - m{x_1} + m = {x_1} + 4\\
x_1^2 - m{x_2} + m = {x_2} + 4
\end{array} \right.\\
\Rightarrow \left( {{x_1} + 4} \right)\left( {{x_2} + 4} \right) = 2\\
\Rightarrow {x_1}{x_2} + 4\left( {{x_1} + {x_2}} \right) + 16 = 2\\
\Rightarrow m - 4 + 4.\left( {m + 1} \right) + 14 = 0\\
\Rightarrow 5m + 14 = 0\\
\Rightarrow m = - \dfrac{{14}}{5}
\end{array}$
