Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=2m\\3x-2y=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}7x=2m+5\\y=m-2x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{2m+5}{7}\\y=\frac{3m-10}{7}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2m+5>0\\3m-10< 0\end{matrix}\right.\) \(\Rightarrow-\frac{5}{2}< m< \frac{10}{3}\)
\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mx-x-my=3m-1\\y=2x-m-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}mx-x-m\left(2x-m-5\right)-3m+1=0\\y=2x-m-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x-mx+m^2+2m+1=0\\y=2x-m-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x\left(m+1\right)+\left(m+1\right)^2=0\\y=2x-m-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+1\\y=m-3\end{matrix}\right.\)
Để hệ PT có nghiệm duy nhất x, y tm x2 + 2y = 0 thì :
⇔ ( m + 1 )2 + 2 ( m - 3 ) = 0
⇔ m2 + 4m - 5 = 0
⇔ ( m - 1 ) ( m + 5 ) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}m-1=0\\m+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=-5\end{matrix}\right.\)
Vậy . . . . . . . . .
\(\left\{{}\begin{matrix}2x+y=m\\3x-2y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=2m\\3x-2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=2m+5\\y=m-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{7}\\y=\dfrac{3m-10}{7}\end{matrix}\right.\)
Để \(x>0;y< 0\Rightarrow\left\{{}\begin{matrix}\dfrac{2m+5}{7}>0\\\dfrac{3m-10}{7}< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m>-\dfrac{5}{2}\\m< \dfrac{10}{3}\end{matrix}\right.\) \(\Rightarrow-\dfrac{5}{2}< m< \dfrac{10}{3}\)