Hệ có nghiệm duy nhất khi \(m^2\ne1\Rightarrow m\ne\pm1\)
Khi đó: \(\left\{{}\begin{matrix}x-my=5-3m\\m^2x-my=2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-my=5-3m\\\left(m^2-1\right)x=5m-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-my=5-3m\\x=\dfrac{5}{m+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{m+1}\\y=\dfrac{3m-2}{m+1}\end{matrix}\right.\)
\(\dfrac{5}{x}+4=\dfrac{3}{y}\Rightarrow m+1+4=\dfrac{3\left(m+1\right)}{3m-2}\) (\(m\ne\dfrac{2}{3}\))
\(\Rightarrow\left(3m-2\right)\left(m+5\right)=3\left(m+1\right)\)
\(\Leftrightarrow3m^2+10m-13=0\Rightarrow\left[{}\begin{matrix}m=1\left(loại\right)\\m=-\dfrac{13}{3}\end{matrix}\right.\)
