Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{-m}{1}\)
=>\(m^2\ne-1\)(luôn đúng)
=>Hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}x-my=1\\mx+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=my+1\\m\left(my+1\right)+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=my+1\\y\left(m^2+1\right)=-m+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-m+3}{m^2+1}\\x=\dfrac{-m^2+3m+m^2+1}{m^2+1}=\dfrac{3m+1}{m^2+1}\end{matrix}\right.\)
\(x^2+y^2-x-3y=0\)
=>\(\left(\dfrac{3m+1}{m^2+1}\right)^2+\left(\dfrac{-m+3}{m^2+1}\right)^2-\dfrac{3m+1}{m^2+1}-\dfrac{3\left(-m+3\right)}{m^2+1}=0\)
=>\(\dfrac{9m^2+6m+1+m^2-6m+9}{\left(m^2+1\right)^2}+\dfrac{-3m-1+3m-9}{m^2+1}=0\)
=>\(\dfrac{10m^2+10}{\left(m^2+1\right)^2}+\dfrac{-10}{m^2+1}=0\)
=>\(\dfrac{10}{m^2+1}-\dfrac{10}{m^2+1}=0\)
=>0m=0(luôn đúng)
=>ĐPCM
