\(I\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\2\left(3-y\right)+my=m+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\6-2y+my=m+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y-3\\my-2y=m-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-3\\y\left(m-2\right)=m-1\end{matrix}\right.\)
Để hpt có nghiệm duy nhất thì \(m-2\ne0\Leftrightarrow m\ne2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-\frac{m+5}{m-1}\\y=\frac{m-1}{m-2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2m-5}{m-2}\\y=\frac{m-1}{m-2}\end{matrix}\right.\)
1)\(x+2y=3\Leftrightarrow\frac{m-1}{m-2}+\frac{4m-10}{m-2}=3\)
\(\Leftrightarrow\frac{5m-11}{m-2}=3\)
Do \(m\ne2\Rightarrow5m-11=3m-6\)
\(\Leftrightarrow2m=5\)
\(\Rightarrow m=\frac{5}{2}\left(t/m\right)\)
2) \(x>1,y< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{m-1}{m-2}>0\left(1\right)\\\frac{2m-5}{m-2}< 0\left(2\right)\end{matrix}\right.\)
Từ 1 : \(\frac{m-1}{m-2}>0\)
TH1:\(\Rightarrow\left\{{}\begin{matrix}m-1>0\\m-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m>1\\m>2\end{matrix}\right.\) \(\Rightarrow m>2\)
TH2 : \(\Leftrightarrow\left\{{}\begin{matrix}m-1< 0\\m-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m< 1\\m< 2\end{matrix}\right.\)\(\Rightarrow m< 1\)
Từ 2 : \(\frac{2m-5}{m-2}< 0\)
TH1: \(\Leftrightarrow\left\{{}\begin{matrix}2m-5>0\\m-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m>\frac{5}{2}\\m< 2\end{matrix}\right.\) \(\Rightarrow\frac{2}{5}< m< 2\)
TH2 \(\Leftrightarrow\left\{{}\begin{matrix}2m-5< 0\\m-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< \frac{5}{2}\\m>2\end{matrix}\right.\)\(\Rightarrow\frac{5}{2}>m>2\)
Vậy để \(x>1,y< 0\) thì
\(\frac{2}{5}< m< 2\) hoặc \(\frac{5}{2}>m>2\)
