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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\left\{{}\begin{matrix}2m+9⋮m+2\\3m+1⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+4+5⋮m+2\\3m+6-5⋮m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5⋮m+2\\-5⋮m+2\end{matrix}\right.\)
=>\(5⋮m+2\)
=>\(m+2\in\left\{1;-1;5;-5\right\}\)
=>\(m\in\left\{-1;-3;3;-7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Để hệ có nghiệm duy nhất thì \(\dfrac{m}{2m}\ne\dfrac{1}{3}\)
=>\(\dfrac{1}{2}\ne\dfrac{1}{3}\)(luôn đúng)
\(\left\{{}\begin{matrix}mx+y=5\\2mx+3y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2mx+2y=10\\2mx+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=4\\mx+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-4\\mx=5-y=5-\left(-4\right)=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-4\\x=\dfrac{9}{m}\end{matrix}\right.\)
\(\left(2m-1\right)\cdot x+\left(m+1\right)\cdot y=m\)
=>\(\dfrac{9}{m}\left(2m-1\right)+\left(m+1\right)\cdot\left(-4\right)=m\)
=>\(\dfrac{9\left(2m-1\right)}{m}=m+4m+4=5m+4\)
=>m(5m+4)=18m-9
=>\(5m^2-14m+9=0\)
=>(m-1)(5m-9)=0
=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{9}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{-2}{-1}=2\)
=>\(m\ne\dfrac{1}{2}\)
\(\left\{{}\begin{matrix}x-2y=5\\mx-y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=5\\y=mx-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2\left(mx-4\right)=5\\y=mx-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(1-2m\right)=5-8=-3\\y=mx-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{2m-1}\\y=\dfrac{3m}{2m-1}-4=\dfrac{3m-4\left(2m-1\right)}{2m-1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{2m-1}\\y=\dfrac{-5m+4}{2m-1}\end{matrix}\right.\)
Để x,y trái dấu thì xy<0
=>\(\dfrac{3\left(-5m+4\right)}{\left(2m-1\right)^2}< 0\)
=>-5m+4<0
=>-5m<-4
=>\(m>\dfrac{4}{5}\)
2: Để x=|y| thì \(\dfrac{3}{2m-1}=\left|\dfrac{-5m+4}{2m-1}\right|\)
=>\(\left[{}\begin{matrix}\dfrac{-5m+4}{2m-1}=\dfrac{3}{2m-1}\\\dfrac{-5m+4}{2m-1}=\dfrac{-3}{2m-1}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-5m+4=3\\-5m+4=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1}{5}\left(nhận\right)\\m=\dfrac{7}{5}\left(nhận\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(m.1+2.7=18\Rightarrow m=4\)
\(\left\{{}\begin{matrix}mx+4y=9\\x+my=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2x+4my=9m\\4x+4my=32\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=9m-32\\mx+4y=9\end{matrix}\right.\)
Hệ có nghiệm duy nhất khi \(m^2-4\ne0\Rightarrow m\ne\pm2\)
Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{9m-32}{m^2-4}\\y=\dfrac{9-mx}{4}=\dfrac{8m-9}{m^2-4}\end{matrix}\right.\)
\(x=3y\Rightarrow\dfrac{9m-32}{m^2-4}=\dfrac{3\left(8m-9\right)}{m^2-4}\)
\(\Rightarrow9m-32=3\left(8m-9\right)\)
\(\Rightarrow m=-\dfrac{1}{3}\)