Để pt có 2 nghiệm khác 0 \(\Leftrightarrow-2m-1\ne0\Rightarrow m\ne-\frac{1}{2}\)
\(a-b+c=1+2m-2m-1=0\)
\(\Rightarrow\) Pt đã cho luôn có 2 nghiệm: \(\left[{}\begin{matrix}x=-1\\x=2m+1\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x_1=-1\\x_2=2m+1\end{matrix}\right.\)
\(\frac{1}{x_1}-\frac{1}{x_2}=3\Leftrightarrow\frac{1}{-1}-\frac{1}{2m+1}=3\)
\(\Leftrightarrow-\frac{1}{2m+1}=4\Rightarrow2m+1=-\frac{1}{4}\Rightarrow m=-\frac{5}{8}\)
TH2: \(\left\{{}\begin{matrix}x_1=2m+1\\x_2=-1\end{matrix}\right.\)
\(\frac{1}{x_1}-\frac{1}{x_2}=3\Leftrightarrow\frac{1}{2m+1}-\frac{1}{-1}=3\)
\(\Leftrightarrow\frac{1}{2m+1}=2\Rightarrow2m+1=\frac{1}{2}\Rightarrow m=-\frac{1}{4}\)
