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

a, \(\left\{{}\begin{matrix}m^2x-my=2m\\x+my=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)x=2m+1\\y=\dfrac{1-x}{m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+1}{m^2+1}\\y=\dfrac{1-\dfrac{2m+1}{m^2+1}}{m}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+1}{m^2+1}\\y=\dfrac{\dfrac{m^2+1-2m-1}{m^2+1}}{m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+1}{m^2+1}\\y=\dfrac{\dfrac{m^2-2m}{m^2+1}}{m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+1}{m^2}\\y=\dfrac{m^2-2m}{m^2+1}:m=\dfrac{m\left(m-2\right)}{m\left(m^2+1\right)}=\dfrac{m-2}{m^2+1}\end{matrix}\right.\)

b, Để hpt có nghiệm duy nhất khi \(\dfrac{m}{1}\ne-\dfrac{1}{m}\Leftrightarrow m^2\ne-1\left(luondung\right)\)

\(\dfrac{2m+1}{m^2}+\dfrac{m-2}{m^2+1}=-1\)

\(\Leftrightarrow\left(2m+1\right)\left(m^2+1\right)+m^2\left(m-2\right)=-m^2\left(m^2+1\right)\)

\(\Leftrightarrow2m^3+2m+m^2+1+m^3-2m^2=-m^4-m^2\)

\(\Leftrightarrow3m^3-m^2+2m+1=-m^4-m^2\)

\(\Leftrightarrow m^4+3m^3+2m+1=0\)

bạn tự giải nhé 

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

=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{9}{5}\end{matrix}\right.\)

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 

TH1: x>0

Hệ phương trình sẽ trở thành \(\left\{{}\begin{matrix}2x-y=1\\mx+y=m+1\end{matrix}\right.\)

Để hệ có nghiệm duy nhất thì \(\dfrac{2}{m}\ne-\dfrac{1}{1}=-1\)

=>\(m\ne-2\)

TH2: x<0

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}-2x-y=1\\mx+y=m+1\end{matrix}\right.\)

Hệ phương trình có nghiệm duy nhất khi \(-\dfrac{2}{m}\ne-\dfrac{1}{1}=-1\)

=>m<>2

Để hệ pt có một nghiệm duy nhất thì \(\dfrac{m}{4}\ne\dfrac{1}{m}\Leftrightarrow2m\ne4\Leftrightarrow m\ne2\)

 Từ pt 1 ta có: y=mx-1 thế vào pt 2 ta đc:

    4x-m(mx-1)=2

  \(\Leftrightarrow4x-m^2x+m=2\)

 \(\Leftrightarrow\left(4-m^2\right)x=2-m\)    (*)

Để hệ có nghiệm duy nhất thì pt (*) phải óc nghiệm duy nhất

    tức \(4-m^2\ne0\Leftrightarrow m^2\ne4\Leftrightarrow m\ne\pm2\)

Vây \(m\ne\pm2\) thì hệ có nghiệm duy nhất

Ta có: \(\left\{{}\begin{matrix}x+my=2\\mx-2y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\m\left(2-my\right)-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\2m-m^2y-2y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\2m-\left(m^2y+2y\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\m^2y+2y=2m-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\y\left(m^2+2\right)=2m-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-my\\y=\dfrac{2m-1}{m^2+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2-\dfrac{m\cdot\left(2m-1\right)}{m^2+2}\\y=\dfrac{2m-1}{m^2+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m^2+4-2m^2+m}{m^2+2}=\dfrac{m+4}{m^2+2}\\y=\dfrac{2m-1}{m^2+2}\end{matrix}\right.\)

Tới đây bạn tự làm tiếp nhé

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)

=>\(m^2\ne1\)

=>\(m\notin\left\{1;-1\right\}\)

\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}mx+m^2y=m^2+m\\mx+y=3m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y\left(m^2-1\right)=m^2+m-3m+1\\x+my=m+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{m^2-2m+1}{\left(m-1\right)\left(m+1\right)}=\dfrac{\left(m-1\right)^2}{\left(m-1\right)\cdot\left(m+1\right)}=\dfrac{m-1}{m+1}\\x=m+1-my\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{m-1}{m+1}\\x=m+1-\dfrac{m^2-m}{m+1}=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\end{matrix}\right.\)

Để x,y đều là số nguyên thì \(\left\{{}\begin{matrix}m-1⋮m+1\\3m+1⋮m+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m+1-2⋮m+1\\3m+3-2⋮m+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-2⋮m+1\\-2⋮m+1\end{matrix}\right.\)

=>\(m+1\in\left\{1;-1;2;-2\right\}\)

=>\(m\in\left\{0;-2;1;-3\right\}\)

mà \(m\notin\left\{1;-1\right\}\)

nên \(m\in\left\{0;-2;-3\right\}\)

\(a,\left\{{}\begin{matrix}mx-y=2m\\x-my=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=2m^2\\x-my=m+1\end{matrix}\right.\)

\(\Leftrightarrow m^2x-x=2m^2-m-1\Leftrightarrow x\left(m^2-1\right)=2m^2-m-1\)

\(ycầuđềbài\Leftrightarrow m^2-1\ne0\Leftrightarrow m\ne\pm-1\)

\(b,\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2m^2-m-1}{m^2-1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{m^2-1}=\dfrac{2m+1}{m+1}=2+\dfrac{-2}{m+1}\\y=mx-2m=\dfrac{m\left(2m+1\right)-2m^2-2m}{m+1}=\dfrac{-m}{m+1}=-1+\dfrac{1}{m+1}\end{matrix}\right.\)

\(\left(x;y\right)\in Z\Leftrightarrow\left\{{}\begin{matrix}m\ne\pm1\\m+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\\m+1\inƯ\left(1\right)=\left\{1;-1\right\}\end{matrix}\right.\)

\(\Rightarrow m=0;m=-2\)