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\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=2m\\3x-2y=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}7x=2m+5\\y=m-2x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{2m+5}{7}\\y=\frac{3m-10}{7}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2m+5>0\\3m-10< 0\end{matrix}\right.\) \(\Rightarrow-\frac{5}{2}< m< \frac{10}{3}\)
Hệ có nghiệm duy nhất khi: \(\dfrac{3}{1}\ne\dfrac{m}{-2}\Rightarrow m\ne-6\)
Khi đó ta có:
\(\left\{{}\begin{matrix}3x+my=5\\x-2y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x+2my=10\\mx-2my=3m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+6\right)x=3m+10\\y=\dfrac{x-3}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{x-3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{-4}{m+6}\end{matrix}\right.\)
\(2x+y=1\Rightarrow\dfrac{2\left(3m+10\right)}{m+6}+\dfrac{-4}{m+6}=1\)
\(\Leftrightarrow\dfrac{6m+16}{m+6}=1\)
\(\Rightarrow6m+16=m+6\)
\(\Rightarrow m=-2\)
Để hệ có nghiệm duy nhất thì \(\dfrac{2}{m}\ne\dfrac{-1}{1}=-1\)
=>\(m\ne-2\)
\(\left\{{}\begin{matrix}2x-y=1\\mx+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y+mx+y=6\\2x-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m+2\right)=6\\y=2x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6}{m+2}\\y=2\cdot\dfrac{6}{m+2}-1=\dfrac{12}{m+2}-1=\dfrac{12-m-2}{m+2}=\dfrac{-m+10}{m+2}\end{matrix}\right.\)
Để x>0 và y<0 thì \(\left\{{}\begin{matrix}\dfrac{6}{m+2}>0\\\dfrac{-m+10}{m+2}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m+2>0\\\dfrac{m-10}{m+2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-2\\\left[{}\begin{matrix}m>10\\m< -2\end{matrix}\right.\end{matrix}\right.\)
=>m>10
Kết hợp điều kiện đề bài và pt thứ 2 của hệ ta được:
\(\left\{{}\begin{matrix}x-y=-6\\2x+y=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=7\end{matrix}\right.\)
Thế vào pt đầu:
\(m.1+2.7=18\Rightarrow m=4\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=m-y\\m-y+ym+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m-y\\ym=1-m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=m-\dfrac{1-m}{m}=\dfrac{m^2+m-1}{m}\\y=\dfrac{1-m}{m}\end{matrix}\right.\)
\(x+2y>0\\ \Leftrightarrow\dfrac{m^2+m-1}{m}+\dfrac{2-2m}{m}>0\\ \Leftrightarrow\dfrac{m^2-m+1}{m}>0\)
Mà \(m^2-m+1=\left(m-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
Vậy \(m>0\) thỏa đề
\(\left\{{}\begin{matrix}2x+y=m\\3x-2y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=2m\\3x-2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=2m+5\\y=m-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{7}\\y=\dfrac{3m-10}{7}\end{matrix}\right.\)
Để \(x>0;y< 0\Rightarrow\left\{{}\begin{matrix}\dfrac{2m+5}{7}>0\\\dfrac{3m-10}{7}< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m>-\dfrac{5}{2}\\m< \dfrac{10}{3}\end{matrix}\right.\) \(\Rightarrow-\dfrac{5}{2}< m< \dfrac{10}{3}\)