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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left\{{}\begin{matrix}mx-y=2m\\x-my=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=2m^2\\x-my=m+1\end{matrix}\right.\)
\(\Leftrightarrow m^2x-x=2m^2-m-1\Leftrightarrow x\left(m^2-1\right)=2m^2-m-1\)
\(ycầuđềbài\Leftrightarrow m^2-1\ne0\Leftrightarrow m\ne\pm-1\)
\(b,\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2m^2-m-1}{m^2-1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{m^2-1}=\dfrac{2m+1}{m+1}=2+\dfrac{-2}{m+1}\\y=mx-2m=\dfrac{m\left(2m+1\right)-2m^2-2m}{m+1}=\dfrac{-m}{m+1}=-1+\dfrac{1}{m+1}\end{matrix}\right.\)
\(\left(x;y\right)\in Z\Leftrightarrow\left\{{}\begin{matrix}m\ne\pm1\\m+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\\m+1\inƯ\left(1\right)=\left\{1;-1\right\}\end{matrix}\right.\)
\(\Rightarrow m=0;m=-2\)
![](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)
\(\left\{{}\begin{matrix}2mx+y=1\\2x-\left(2m+1\right)y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\left(2m+1\right)y+y=1\\2x=\left(2m+1\right)y-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2m^2y+my+y-1=0\\2x=\left(2m+1\right)y-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\left(2m^2+m+1\right)=1\left(1\right)\\2x=\left(2m+1\right)y-1\end{matrix}\right.\)
Để pt có nghiệm duy nhất tức là pt (1) có nghiệm duy nhất
\(\Leftrightarrow2m^2+m+1\ne0\Leftrightarrow m^2+\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ne0\) ( luôn đúng )
Vậy với mọi giá trị m thỏa mãn là pt có nghiệm duy nhất.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}mx+y=2m\\x+my=m+1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=2m-mx\\x+m\left(2m-mx\right)=m+1\left(1\right)\end{matrix}\right.\)
(1) ⇔x+2m2-m2x=m+1
⇔x(1-m2)=m+1-2m2
TH1: 1-m2=0
⇔m=\(\pm\)1
-Thay m= 1 vào (2) ta có: 0x =0 (luôn đúng)
⇒m=1(chọn)
-Thay m=-1 và (2) ta có: 0x=-2 (vô lí)
⇒m=-1(loại)
TH2: 1-m2 ≠ 0
⇔m ≠ \(\pm\) 1
⇒HPT có nghiệm duy nhất:
x= \(\dfrac{-2m^2+m+1}{1-m^2}\)
y= \(2m-m.\dfrac{-2m^2+m+1}{1-m^2}\)
⇔y= \(2m+\dfrac{-2m^3-m^2-m}{1-m^2}\)
\(\left\{{}\begin{matrix}\left(2m+1\right)x+y=2m-2\left(1\right)\\m^2x-y=m^2-3m\end{matrix}\right.\)
\(\Rightarrow\left(m^2+2m+1\right)x=m^2-m-2\)
\(\Rightarrow x=\dfrac{m^2-m-2}{m^2+2m+1}\left(m\ne-1\right)\)
\(\Rightarrow x=1+\dfrac{-3m-3}{m^2+2m+1}=1+\dfrac{-3\left(m+1\right)}{\left(m+1\right)^2}=1+\dfrac{-3}{m+1}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow y=2m-2-\left(2m+1\right)\left(1-\dfrac{3}{m+1}\right)\)
\(\Rightarrow y=\dfrac{3m}{m+1}=3+\dfrac{-1}{m+1}\)
\(\Rightarrow x,y\in Z\left(m\in Z\right)\Leftrightarrow\left\{{}\begin{matrix}m+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\\m+1\inƯ\left(1\right)=\left\{\pm1\right\}\end{matrix}\right.\)
\(\Rightarrow m+1=\pm1\Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=-2\left(tm\right)\end{matrix}\right.\)