a: Để hệ có nghiệm duy nhất thì \(\dfrac{3}{1}\ne\dfrac{m}{1}\)
=>\(m\ne3\)
b:Khi m<>3 thì \(\left\{{}\begin{matrix}3x+my=4\\x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+my=4\\3x+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\left(m-3\right)=1\\x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1}{m-3}\\x=1-\dfrac{1}{m-3}=\dfrac{m-3-1}{m-3}=\dfrac{m-4}{m-3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< 0\\y>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{m-3}>0\\\dfrac{m-4}{m-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m-3>0\\3< m< 4\end{matrix}\right.\)
=>3<m<4
giải câu d thôi ạ
