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\(\Rightarrow\left\{{}\begin{matrix}27x=m+3\\25x-3y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{m+3}{27}\\y=\frac{25x-3}{3}=\frac{25m-6}{81}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{m+3}{27}>0\\\frac{25m-6}{81}< 0\end{matrix}\right.\) \(\Rightarrow-3< m< \frac{6}{25}\)
\(\left\{{}\begin{matrix}2x+3y=m\\25x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-m=3y\\25x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{m-2x}{3}\\25x+3x-m=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3-2x}{3}\\27x=3+m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3+m}{27}\\y=\frac{m-\frac{6+2m}{27}}{3}=\frac{27m-6-2m}{81}\end{matrix}\right.\)
Mà: \(x>0;y< 0\Rightarrow\left\{{}\begin{matrix}\frac{3+m}{27}>0\\\frac{25m-6}{81}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-3\\m< \frac{6}{25}\end{matrix}\right.\)
\(\Rightarrow m\in\left\{-3;\frac{6}{25}\right\}\)
\(\left\{{}\begin{matrix}6x-3my=-9\\m^2x+3my=4m\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}6x-3my=-9\\\left(m^2+6\right)x=4m-9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{4m-9}{m^2+6}\\y=\frac{3m+8}{m^2+6}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< 0\\y>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{4m-9}{m^2+6}< 0\\\frac{3m+8}{m^2+6}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m< \frac{9}{4}\\m>-\frac{8}{3}\end{matrix}\right.\)
\(\Rightarrow-\frac{8}{3}< m< \frac{9}{4}\)
Ta có: \(\left\{{}\begin{matrix}2x+3y=m\\5x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=m\\15x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x=m+3\\5x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{17}\\y=5x-1=\dfrac{5m+15}{17}-\dfrac{17}{17}=\dfrac{5m-2}{17}\end{matrix}\right.\)
Để hệ phương trình có nghiệm duy nhất sao cho x<0 và y>0 thì
\(\left\{{}\begin{matrix}\dfrac{m+3}{17}< 0\\\dfrac{5m-2}{17}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+3< 0\\5m-2>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< -3\\m>\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow m\in\varnothing\)
=>x=2m+3y và 2(2m+3y)-5y=m+1
=>x=2m+3y và 4m+6y-5y-m-1=0
=>x=2m+3y và y+3m-1=0
=>y=-3m+1 và x=2m+3(-3m+1)=2m-9m+3=-7m+3
Để x>0; y<0 thì -7m+3>0 và -3m+1<0
=>-7m>-3 và -3m<-1
=>m<3/7 và m>1/3
1)
\(\left\{{}\begin{matrix}x+y=4\\2x+3y=m\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+3y=12\\2x+3y=m\end{matrix}\right.\)
trừ 2 vế của pt cho nhau ta tìm được
\(\left\{{}\begin{matrix}x=12-m\\y=m-8\end{matrix}\right.\)
để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m< 12\\m< 8\end{matrix}\right.\Rightarrow}m< 8}\)
x=\(\dfrac{m-3y}{2}\)
=> \(25.\dfrac{m-3y}{2}-3y=3\)
=> 25(m-3y)-6y=6
=> 25m-75y-6y-6=0
=>25m-81y-6=0
=>25m-6=81y
=>y=\(\dfrac{25m-6}{81}\)
=>x=\(\dfrac{m-1}{27}\)
voi x>0 thi \(\dfrac{m-1}{27}>0\)
=> m-1>0
=> m>1
voi y<0 thi \(\dfrac{25m-6}{81}< 0\)
=> 25m-6<0
=> m<6/25
Sao x= \(\dfrac{m-1}{27}\) đấy