Question

Answer 1

Để hệ có nghiệm duy nhất thì \(\dfrac{m}{1}\ne-\dfrac{1}{m}\)

=>\(m^2\ne-1\)(luôn đúng)

=>Hệ luôn có nghiệm duy nhất

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

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

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

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

3x-y=1

=>\(\dfrac{3\left(2m+6\right)-m^2-6m+1}{m^2+1}=1\)

=>\(6m+18-m^2-6m+1=m^2+1\)

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

=>\(m^2=9\)

=>\(\left[{}\begin{matrix}m=3\\m=-3\end{matrix}\right.\)