Đáp án đúng: B
Giải chi tiết:Phương trình \(\left( {m + 1} \right){x^2} - 2\left( {m - 1} \right)x + m - 2 = 0\) có hai nghiệm \({x_1},\,{x_2}\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}m + 1 \ne 0\\\Delta ' \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne - 1\\{\left( {m - 1} \right)^2} - \left( {m + 1} \right)\left( {m - 2} \right) \ge 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m \ne - 1\\{m^2} - 2m + 1 - {m^2} + m + 2 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne - 1\\ - m + 3 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ne - 1\\m \le 3\end{array} \right.\,\,\end{array}\)
Theo hệ thức Vi ét ta có : \(\left\{ \begin{array}{l}{x_1} + {x_2} = \frac{{2\left( {m - 1} \right)}}{{m + 1}}\,\,\,\,\left( 1 \right)\\{x_1}{x_2} = \frac{{m - 2}}{{m + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\)
Theo đề bài ta có :
\(\begin{array}{l}\left| {{x_1} + {x_2}} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}{x_1} + {x_2} = 2\\{x_1} + {x_2} = - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\frac{{2\left( {m - 1} \right)}}{{m + 1}} = 2\\\frac{{2\left( {m - 1} \right)}}{{m + 1}} = - 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2m - 2 = 2m + 2\\2m - 2 = - 2m - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 2 = 2\,\,\,\left( {ktm} \right)\\4m = 0\end{array} \right. \Leftrightarrow m = 0\,\,\,\,\left( {tm} \right)\end{array}\)
Chọn B.
