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: 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 

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

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

(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= \(2m+\dfrac{-2m^3-m^2-m}{1-m^2}\)

• PT có nghiệm duy nhất \( \Leftrightarrow \dfrac{1}{m} \ne \dfrac{-2}{1} \Leftrightarrow m \ne \dfrac{-1}{2}\)

• PT vô nghiệm \(\Leftrightarrow \dfrac{1}{m} =\dfrac{-2}{1}  \ne \dfrac{1}{2} \Leftrightarrow m=\dfrac{-1}{2}\)

• PT có vô số nghiệm \(\Leftrightarrow \dfrac{1}{m} = \dfrac{-2}{1} = \dfrac{1}{2} (\text{Vô lý})\)

Vậy....

a, Thay \(m=-1\) vào

\(=>\left\{{}\begin{matrix}-x+y=1\\2x-y=-1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

b, Để hệ pt có nghiệm duy nhất :

\(\dfrac{m}{2}\ne\dfrac{1}{-1}\\ =>\dfrac{m}{2}\ne-1\\ =>m\ne-2\)

a.

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\3x=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=2-\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)

b.

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

Phương trình có nghiệm khi \(m\ne1\)

Khi đó: \(x=\dfrac{m-2}{m-1}=1-\dfrac{1}{m-1}\)

\(x\in Z\Rightarrow\dfrac{1}{m-1}\in Z\Rightarrow m-1=Ư\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow m=\left\{0;2\right\}\)

- Không có đề à bạn

Thay \(a=-\sqrt{2}\) vào pt :

\(\left\{{}\begin{matrix}\left(-\sqrt{2}+1\right)x-y=3\left(1\right)\\-\sqrt{2}x+y=-\sqrt{2}\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)+\left(2\right):\) 

\(\left(-\sqrt{2}+1-\sqrt{2}\right)x=3-\sqrt{2}\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{2}}{1-2\sqrt{2}}\)

\(\Leftrightarrow x=\dfrac{1-5\sqrt{2}}{7}\)\(\left(3\right)\)

Thay \(\left(3\right)\) vào \(\left(2\right)\) : \(-\sqrt{2}.\dfrac{1-5\sqrt{2}}{7}+y=-\sqrt{2}\)

\(\Rightarrow y=\)\(-\sqrt{2}+\dfrac{6\sqrt{2}}{7}\)

\(\Rightarrow y=-\dfrac{\sqrt{2}}{7}\)

Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(\dfrac{1-5\sqrt{2}}{7};-\dfrac{\sqrt{2}}{7}\right)\)