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\(\left\{{}\begin{matrix}2x+3y=m\\25x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-m=3y\\25x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{m-2x}{3}\\25x+3x-m=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3-2x}{3}\\27x=3+m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3+m}{27}\\y=\frac{m-\frac{6+2m}{27}}{3}=\frac{27m-6-2m}{81}\end{matrix}\right.\)
Mà: \(x>0;y< 0\Rightarrow\left\{{}\begin{matrix}\frac{3+m}{27}>0\\\frac{25m-6}{81}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-3\\m< \frac{6}{25}\end{matrix}\right.\)
\(\Rightarrow m\in\left\{-3;\frac{6}{25}\right\}\)
\(\left\{{}\begin{matrix}3x+2y=10\\2x-y=m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=10\\4x-2y=2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=10+2m\\3x+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10+2m}{7}\\3\left(\dfrac{10+2m}{7}\right)+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10+2m}{7}\\\dfrac{30+6m}{7}+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10+2m}{7}\\y=\dfrac{40-6m}{14}\end{matrix}\right.\)
Để \(x>0\) \(\Leftrightarrow\dfrac{10+2m}{7}>0\)
\(\Leftrightarrow m>-5\) (1)
Để \(y>0\) \(\Leftrightarrow40-6m< 0\)
\(\Leftrightarrow m>\dfrac{20}{3}\) (2)
\(\left(1\right);\left(2\right)\rightarrow m>\dfrac{20}{3}\)
Vậy \(m>\dfrac{20}{3}\) thì \(x>0;y< 0\)
\(\Rightarrow\left\{{}\begin{matrix}27x=m+3\\25x-3y=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{m+3}{27}\\y=\frac{25x-3}{3}=\frac{25m-6}{81}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{m+3}{27}>0\\\frac{25m-6}{81}< 0\end{matrix}\right.\) \(\Rightarrow-3< m< \frac{6}{25}\)
Lời giải:
Từ PT$(1)\Rightarrow x=m+1-my$. Thay vô PT(2):
$m(m+1-my)+y=3m-1$
$\Leftrightarrow y(1-m^2)+m^2+m=3m-1$
$\Leftrightarrow y(1-m^2)=-m^2+2m-1(*)$
Để hpt có nghiệm $(x,y)$ duy nhất thì pt $(*)$ cũng phải có nghiệm $y$ duy nhất
Điều này xảy ra khi $1-m^2\neq 0\Leftrightarrow m\neq \pm 1$
Khi đó: $y=\frac{-m^2+2m-1}{1-m^2}=\frac{-(m-1)^2}{-(m-1)(m+1)}=\frac{m-1}{m+1}$
$x=m+1-my=m+1-\frac{m(m-1)}{m+1}=\frac{3m+1}{m+1}$
Có:
$x+y=\frac{m-1}{m+1}+\frac{3m+1}{m+1}=\frac{4m}{m+1}<0$
$\Leftrightarrow -1< m< 0$
Kết hợp với đk $m\neq \pm 1$ suy ra $-1< m< 0$ thì thỏa đề.
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
\(\left\{{}\begin{matrix}\dfrac{m+3}{17}< 0\\\dfrac{5m-2}{17}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+3< 0\\5m-2>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< -3\\m>\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow m\in\varnothing\)