Question

Tìm m để ​\(\left\{{}\begin{matrix}mx-y=1\\x+my=m+6\end{matrix}\right.\)​ có nhiệm thỏa mãn 3x-y=1

Answer 1

HPT \(\Leftrightarrow\left\{{}\begin{matrix}y=mx-1\\x+m^2x-m=m+6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=mx-1\\\left(1+m^2\right)x=2m+6\end{matrix}\right.\)

HPT có 2 nghiệm vì \(m^2+1>0\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{2m+6}{m^2+1}\\y=\frac{m^2+6m-1}{m^2+1}\end{matrix}\right.\)

Ta có: \(3x-y=1\)

\(\Rightarrow\frac{6m+18}{m^2+1}-\frac{m^2+6m-1}{m^2+1}=1\)

\(\Rightarrow-m^2+19=m^2+1\)

\(\Rightarrow m=\pm3\)

Vậy \(m=\pm3\)