Đáp án: $\forall m \ne 0;m \ne 2\,$
Giải thích các bước giải:
$\begin{array}{l}
pt\,co\,2\,nghiem\,phan\,biet\,{x_1},{x_2}\\
\Leftrightarrow m \ne 0;\Delta > 0\\
\Rightarrow m \ne 0;{\left( {m - 2} \right)^2} - 4.m.\left( { - 2} \right) > 0\\
\Rightarrow m \ne 0;{m^2} - 4m + 4 + 8m > 0\\
\Rightarrow m \ne 0;{m^2} + 4m + 4 > 0\\
\Rightarrow m \ne 0;{\left( {m + 2} \right)^2} > 0\\
\Rightarrow m \ne - 2;m \ne 0\\
+ ){x_1} < {x_2} \le - 1\\
\Rightarrow \left( {{x_1} + 1} \right)\left( {{x_2} + 1} \right) \ge 0\\
\Rightarrow {x_1}{x_2} + {x_1} + {x_2} + 1 \ge 0\\
\Rightarrow \frac{{ - 2}}{m} + \frac{{ - (m - 2)}}{m} + 1 \ge 0\\
\Rightarrow \frac{{ - m}}{m} + 1 \ge 0 \Rightarrow 0 \ge 0\,\left( {luon\,dung} \right)\\
Vay\forall m \ne 0;m \ne 2\,thi\,tmyc
\end{array}$
