Question

Cho hệ \(\left\{{}\begin{matrix}x-my=2-4m\\mx+y=3m+1\end{matrix}\right.\)

Gọi \(\left(x_0:y_0\right)\) là nghiệm duy nhất của hệ

CMR: \(x_0^2+y_0^2-5\left(x_0+y_0\right)+10=0\)

Answer 1

\(\left\{{}\begin{matrix}x_0-my_0=2-4m\\mx_0+y_0=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0-2=m\left(y_0-4\right)\\y_0-1=m\left(3-x_0\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x_0-2\right)\left(3-x_0\right)=m\left(y_0-4\right)\left(3-x_0\right)\\\left(y_0-1\right)\left(y_0-4\right)=m\left(y_0-4\right)\left(3-x_0\right)\end{matrix}\right.\)

\(\Rightarrow\left(x_0-2\right)\left(3-x_0\right)=\left(y_0-1\right)\left(y_0-4\right)\)