Question

\(\left\{{}\begin{matrix}mx+4y=9\\x+my=8\end{matrix}\right.\)

tìm m để hệ có nghiệm duy nhất thỏa mãn `x=3y`

Answer 1

\(\left\{{}\begin{matrix}mx+4y=9\\x+my=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2x+4my=9m\\4x+4my=32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=9m-32\\mx+4y=9\end{matrix}\right.\) 

Hệ có nghiệm duy nhất khi \(m^2-4\ne0\Rightarrow m\ne\pm2\)

Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{9m-32}{m^2-4}\\y=\dfrac{9-mx}{4}=\dfrac{8m-9}{m^2-4}\end{matrix}\right.\)

\(x=3y\Rightarrow\dfrac{9m-32}{m^2-4}=\dfrac{3\left(8m-9\right)}{m^2-4}\)

\(\Rightarrow9m-32=3\left(8m-9\right)\)

\(\Rightarrow m=-\dfrac{1}{3}\)