a/ câu a xem lại đề bài
b/ \(\Delta'=\left(m+1\right)^2-m^2-1=m^2+2m+1-m^2-1=2m\)
Để PT có 2 no\(\Leftrightarrow2m\ge0\Leftrightarrow m\ge0\)
Theo Vi-ét có:
\(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=m^2+1\end{matrix}\right.\)
Có \(\frac{x_1}{x_2}+\frac{x_2}{x_1}=4\)
\(\Leftrightarrow x_1^2+x_2^2=4x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-6x_1x_2=0\)
\(\Leftrightarrow\left(2m-2\right)^2-6\left(m^2+1\right)=0\)
\(\Leftrightarrow4m^2-8m+4-6m^2-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-2+\sqrt{3}\\m=-2-\sqrt{3}\end{matrix}\right.\) (loại)