Đáp án:
\(m \ne 0\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
⇔Δ>0
\(\begin{array}{l}
\to {m^2} + 4m + 4 - 8m > 0\\
\to {m^2} - 4m + 4 > 0\\
\to {\left( {m - 2} \right)^2} > 0\\
\Leftrightarrow m \ne 2\\
Có: - 1 < \left| {\frac{{2\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}}} \right| < 1\\
Do: - 1 < \left| {\frac{{2\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}}} \right|\left( {ld} \right)\\
\to Xet:\left| {\frac{{2\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}}} \right| < 1\\
\to \left[ \begin{array}{l}
\frac{{2\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}} < 1\\
\frac{{2\left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}} > - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{2\left( {m + 2} \right)}}{{2m}} < 1\\
\frac{{2\left( {m + 2} \right)}}{{2m}} > - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{m + 2 - m}}{m} < 0\\
\frac{{m + 2 + m}}{m} > 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{2}{m} < 0\\
\frac{{2m + 2}}{m} > 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m < 0\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
2\left( {m + 1} \right) > 0\\
m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2\left( {m + 1} \right) < 0\\
m < 0
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
m < 0\\
m > 0\\
m < - 1
\end{array} \right.\\
KL:m \ne 0
\end{array}\)
