Với \(m\ne0\) \(\left\{{}\begin{matrix}x_1+x_2=\frac{2m}{m-2}\\x_1x_2=\frac{m-4}{m-2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\left(x_1+x_2\right)-2\left(x_1+x_2\right)=2m\\mx_1x_2-2x_1x_2=m-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\left(x_1+x_2-2\right)=2\left(x_1+x_2\right)\\m\left(x_1x_2-1\right)=2x_1x_2-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m=\frac{2\left(x_1+x_2\right)}{x_1+x_2-2}\\m=\frac{2x_1x_2-4}{x_1x_2-1}\end{matrix}\right.\)
\(\Rightarrow\frac{2\left(x_1+x_2\right)}{x_1+x_2-2}=\frac{2x_1x_2-4}{x_1x_2-1}\)
b/ \(\left\{{}\begin{matrix}y_1=-x_1\\y_2=-x_2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=-\left(x_1+x_2\right)=\frac{-2m}{\left(m-2\right)}\\y_1y_2=x_1x_2=\frac{m-4}{m-2}\end{matrix}\right.\)
Theo Viet đảo, \(y_1;y_2\) là nghiệm:
\(y^2+\frac{2m}{m-2}y+\frac{m-4}{m-2}=0\Leftrightarrow\left(m-2\right)y^2+2my+m-4=0\) \(\left(m\ne2\right)\)
