a) Ta có
\(\Delta'=\left[-\left(m-1\right)\right]^2-\left(m^2-1\right)=m^2-2m+1-m^2+1=2-2m=2\left(1-m\right)\) Để phương trình có hai nghiệm
\(\Leftrightarrow\Delta\ge0\Leftrightarrow1-m\ge0\Leftrightarrow m\le1\)
Theo Vi-ét
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m^2-1\end{matrix}\right.\)
Ta có
\(\frac{1}{x_1}+\frac{1}{x_2}=\frac{1}{2}\Leftrightarrow\frac{x_1+x_2}{x_1x_2}=\frac{1}{2}\Leftrightarrow\frac{2\left(m-1\right)}{m^2-1}=\frac{1}{2}\Leftrightarrow4m-4=m^2-1\)
\(\Leftrightarrow m^2-1-4m+4=0\Leftrightarrow m^2-4m+4-1=0\Rightarrow\left(m-2\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}m-2=1\\m-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=3\left(ktm\right)\\m=1\left(tm\right)\end{matrix}\right.\)
Vậy m = 1
