Đáp án:
$\begin{array}{l}
a)x = 5\\
\Rightarrow {5^2} - m.5 - m - 1 = 0\\
\Rightarrow 25 - 5m - m - 1 = 0\\
\Rightarrow 6m = 24\\
\Rightarrow m = 4\\
\Rightarrow {x^2} - 4x - 5 = 0\\
\Rightarrow \left( {x - 5} \right)\left( {x + 1} \right) = 0\\
\Rightarrow x = 5;x = - 1\\
b){x^2} - mx - m - 1 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta > 0\\
\dfrac{{ - b}}{a} < 0\\
\dfrac{c}{a} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{m^2} + 4m + 4 > 0\\
m < 0\\
- m - 1 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {m + 2} \right)^2} > 0\\
m < 0\\
m < - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 2\\
m < - 1
\end{array} \right.\\
Vậy\,m < - 1;m \ne - 2\\
c)\left\{ \begin{array}{l}
a.c < 0\\
{x_1} + {x_2} < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- m - 1 < 0\\
m < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > - 1\\
m < 0
\end{array} \right.\\
\Rightarrow - 1 < m < 0\\
d)\left\{ \begin{array}{l}
\Delta > 0\\
{x_1}{x_2} > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 2\\
- m - 1 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 2\\
m < - 1
\end{array} \right.\\
Vậy\,m < - 1;m \ne - 2
\end{array}$
