\(\Delta'=\left(m-1\right)^2+m=m^2+m+1=\left(m+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall m\)
\(\Rightarrow\) Phương trình luôn có nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m\end{matrix}\right.\)
Do phương trình ẩn y có nghiệm \(\left\{{}\begin{matrix}y_1=x_1+\frac{1}{x_2}\\y_2=x_2+\frac{1}{x_1}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=x_1+x_2+\frac{1}{x_1}+\frac{1}{x_2}\\y_1y_2=\left(x_1+\frac{1}{x_2}\right)\left(x_2+\frac{1}{x_1}\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=x_1+x_2+\frac{x_1+x_2}{x_1x_2}\\y_1y_2=x_1x_2+\frac{1}{x_1x_2}+2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=2\left(m-1\right)+\frac{2\left(m-1\right)}{-m}\\y_1y_2=-m-\frac{1}{m}+2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=\frac{-m\left(2m-2\right)+2m-2}{-m}=\frac{2\left(m-1\right)^2}{m}\\y_1y_2=\frac{-m^2+2m-1}{m}=-\frac{\left(m-1\right)^2}{m}\end{matrix}\right.\)
Theo Viet đảo, \(y_1\) và \(y_2\) là nghiệm của:
\(y^2-\frac{2\left(m-1\right)^2}{m}y-\frac{\left(m-1\right)^2}{m}=0\)
\(\Leftrightarrow my^2-2\left(m-1\right)^2-\left(m-1\right)^2=0\) (\(m\ne0\))