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1, Gỉa sử m = 1
Thay m = 1 vào hpt trên ta được
\(\left\{{}\begin{matrix}x+y=1\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
2, Để hệ có nghiệm duy nhất \(\dfrac{m}{4}\ne\dfrac{1}{m}\Leftrightarrow m^2\ne4\Leftrightarrow m\ne\pm2\)
\(\left\{{}\begin{matrix}m^2x+my=m\\4x+my=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=m-2\\y=1-mx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{m+2}\\y=1-\dfrac{m}{m+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{m+2}\\y=\dfrac{2}{m+2}\end{matrix}\right.\)
Ta có : \(\dfrac{1}{m+2}-\dfrac{2}{m+2}=1\Rightarrow1-2=m+2\Leftrightarrow-1=m+2\Leftrightarrow m=-3\)(tmđk)
a, Với m = 1
\(\left\{{}\begin{matrix}x+y=1_{\left(1\right)}\\4x+y=2_{\left(2\right)}\end{matrix}\right.\)
Lấy (2) - (1) ta được
\(3x=1\Leftrightarrow x=\dfrac{1}{3};\Rightarrow y=1-x=1-\dfrac{1}{3}=\dfrac{2}{3}\)
Vậy (x,y) = \(\left(\dfrac{1}{3};\dfrac{2}{3}\right)\)
c, no của hệ là
\(\left(\dfrac{-1}{m+2};\dfrac{2m+2}{m+2}\right)\\ Theo.bài:\\ x-y=1\\ \Leftrightarrow\dfrac{-1}{m+2}-\dfrac{2m+2}{m+2}=1\\ \Leftrightarrow-1-2m-2=m+2\\ \Leftrightarrow3m=-5\\ m=\dfrac{-5}{3}\)
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
Khi \(m\notin\left\{1;-1\right\}\) thì \(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y-2m=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-m^2+1\right)=-m^2+m\\x=m+1-my\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m^2-m}{m^2-1}=\dfrac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{m}{m+1}\\x=m+1-\dfrac{m^2}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m}{m+1}\\x=\dfrac{\left(m+1\right)^2-m^2}{m+1}=\dfrac{2m+1}{m+1}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>=2\\y>=1\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}>=2\\\dfrac{m}{m+1}>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2\left(m+1\right)}{m+1}>=0\\\dfrac{m-m-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2m-2}{m+1}>=0\\\dfrac{-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{m+1}>=0\\-\dfrac{1}{m+1}>=0\end{matrix}\right.\Leftrightarrow m+1< 0\)
=>m<-1
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{-2}{-m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
\(\left\{{}\begin{matrix}mx-2y=2m-1\\2x-my=9-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=mx-2m+1\\2x-my=9-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\2x-m\left(x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\right)=9-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\2x-x\cdot\dfrac{m^2}{2}+m^2-\dfrac{1}{2}m=9-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\x\left(2-\dfrac{m^2}{2}\right)=-m^2+\dfrac{1}{2}m-3m+9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\\x\cdot\dfrac{4-m^2}{2}=-m^2-\dfrac{5}{2}m+9=\dfrac{-2m^2-5m+18}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{-2m^2-5m+18}{4-m^2}=\dfrac{2m^2+5m-18}{m^2-4}\\y=x\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{\left(2m+9\right)\left(m-2\right)}{\left(m-2\right)\left(m+2\right)}=\dfrac{2m+9}{m+2}\\y=\dfrac{2m+9}{m+2}\cdot\dfrac{m}{2}-m+\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m+9}{m+2}\\y=\dfrac{2m^2+9m-2m\left(m+2\right)+m+2}{2\left(m+2\right)}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m+9}{m+2}\\y=\dfrac{2m^2+10m+2-2m^2-4m}{2\left(m+2\right)}=\dfrac{6m+2}{2\left(m+2\right)}=\dfrac{3m+1}{m+2}\end{matrix}\right.\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}2m+9⋮m+2\\3m+1⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+4+5⋮m+2\\3m+6-5⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5⋮m+2\\-5⋮m+2\end{matrix}\right.\)
=>\(5⋮m+2\)
=>\(m+2\in\left\{1;-1;5;-5\right\}\)
=>\(m\in\left\{-1;-3;3;-7\right\}\)
Lời giải:
$x+my=2\Rightarrow x=2-my$. Thay vào PT(2):
$m(2-my)-2y=1$
$\Leftrightarrow 2m-y(m^2+2)=1$
$\Leftrightarrow y=\frac{2m-1}{m^2+2}$
$x=2-my=2-\frac{2m^2-m}{m^2+2}=\frac{m+4}{m^2+2}$
Vậy hpt có nghiệm $(x,y)=(\frac{m+4}{m^2+2}; \frac{2m-1}{m^2+2})$
Để $x<0; y>0$
$\Leftrightarrow \frac{m+4}{m^2+2}<0$ và $\frac{2m-1}{m^2+2}>0$
$\Leftrightarrow m+4<0$ và $2m-1>0$ (do $m^2+2>0$)
$\Leftrightarrow m< -4$ và $m> \frac{1}{2}$ (vô lý)
Do đó không tồn tại $m$ thỏa mãn đề.
Hệ có nghiệm duy nhất khi: \(\dfrac{1}{m}\ne\dfrac{m}{-2}\Rightarrow m^2\ne-2\) (luôn đúng)
\(\Rightarrow\) Hệ luôn có nghiệm duy nhất với mọi m
Khi đó: \(\left\{{}\begin{matrix}x+my=2\\mx-2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+2my=4\\m^2x-2my=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+2\right)x=m+4\\y=\dfrac{mx-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+4}{m^2+2}\\y=\dfrac{4m-2}{2\left(m^2+2\right)}\end{matrix}\right.\)
Nghiệm hệ thỏa mãn x<0, y<0 \(\Rightarrow\left\{{}\begin{matrix}\dfrac{m+4}{m^2+2}< 0\\\dfrac{4m-2}{2\left(m^2+2\right)}< 0\end{matrix}\right.\) (1)
Do \(m^2+2>0;\forall m\) nên (1) tương đương:
\(\left\{{}\begin{matrix}m+4< 0\\4m-2< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m< -4\\m< \dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow m< -4\)
a: Khi m=2 thì hệ sẽ là;
2x-y=4 và x-2y=3
=>x=5/3 và y=-2/3
b: mx-y=2m và x-my=m+1
=>x=my+m+1 và m(my+m+1)-y=2m
=>m^2y+m^2+m-y-2m=0
=>y(m^2-1)=-m^2+m
Để phương trình có nghiệm duy nhất thì m^2-1<>0
=>m<>1; m<>-1
=>y=(-m^2+m)/(m^2-1)=(-m)/m+1
x=my+m+1
\(=\dfrac{-m^2+m^2+2m+1}{m+1}=\dfrac{2m+1}{m+1}\)
x^2-y^2=5/2
=>\(\left(\dfrac{2m+1}{m+1}\right)^2-\left(-\dfrac{m}{m+1}\right)^2=\dfrac{5}{2}\)
=>\(\dfrac{4m^2+4m+1-m^2}{\left(m+1\right)^2}=\dfrac{5}{2}\)
=>2(3m^2+4m+1)=5(m^2+2m+1)
=>6m^2+8m+2-5m^2-10m-5=0
=>m^2-2m-3=0
=>(m-3)(m+1)=0
=>m=3
1: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{1}{-1}=-1\)
=>\(m\ne-1\)
2: \(\left\{{}\begin{matrix}x+y=1\\mx-y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y+mx-y=1+2m\\x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m+1\right)=2m+1\\x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m+1}{m+1}\\y=1-x=1-\dfrac{2m+1}{m+1}=\dfrac{m+1-2m-1}{m+1}=-\dfrac{m}{m+1}\end{matrix}\right.\)
x+2y=2
=>\(\dfrac{2m+1}{m+1}+\dfrac{-2m}{m+1}=2\)
=>\(\dfrac{1}{m+1}=2\)
=>\(m+1=\dfrac{1}{2}\)
=>\(m=-\dfrac{1}{2}\left(nhận\right)\)
Đề không hiển thị hệ phương trình bạn à.