Đáp án:
$\begin{array}{l}
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - m - 1\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
\Leftrightarrow {x_1} + {x_2} + {x_1}{x_2} = - m - 1 + m - 1 = - 2\\
Vậy\,{x_1} + {x_2} + {x_1}{x_2} = - 2\\
d)Do:x_1^2 + \left( {m + 1} \right).{x_1} + m - 1 = 0\\
\Leftrightarrow x_1^2 = - \left( {m + 1} \right){x_1} - m + 1\\
Khi:x_1^2 - \left( {m + 1} \right).{x_2} - 2 = 0\\
\Leftrightarrow - \left( {m + 1} \right){x_1} - m + 1 - \left( {m + 1} \right).{x_2} - 2 = 0\\
\Leftrightarrow \left( {m + 1} \right).\left( {{x_1} + {x_2}} \right) + m + 1 = 0\\
\Leftrightarrow \left( {m + 1} \right).\left( { - m - 1} \right) + m + 1 = 0\\
\Leftrightarrow \left( {m + 1} \right).\left( { - m} \right) = 0\\
\Leftrightarrow m = 0/m = - 1
\end{array}$
Vậy m=0 hoặc m=-1
