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

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

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

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

Với m = 1 ta có: \(\left\{{}\begin{matrix}x=2-y\\0y=0\left(VSN\right)\end{matrix}\right.\)

\(\Rightarrow\) Hpt vô số nghiệm

Với m = -1 ta có: \(\left\{{}\begin{matrix}x=y\\0y=4\left(VN\right)\end{matrix}\right.\)

\(\Rightarrow\) Hpt vô nghiệm

Với m \(\ne\) \(\pm\)1 ta có: \(\left\{{}\begin{matrix}x=m+1-my\\y=\dfrac{m^2-2m+1}{m^2-1}\end{matrix}\right.\)

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

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

Vậy hpt có nghiệm duy nhất x = ..; y = ... với x \(\ne\) \(\pm\) 1

Ta có: x = |y|

\(\Leftrightarrow\) \(\dfrac{3m+1}{m+1}=\left|\dfrac{m-1}{m+1}\right|\) 

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\dfrac{3m+1}{m+1}=\dfrac{m-1}{m+1}\\\dfrac{3m+1}{m+1}=\dfrac{1-m}{m+1}\end{matrix}\right.\)

\(\Rightarrow\) \(\left[{}\begin{matrix}3m+1=m-1\\3m+1=1-m\end{matrix}\right.\) (Vì m \(\ne\) -1)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2m=-2\\4m=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=-1\\m=0\end{matrix}\right.\) 

Vì m \(\ne\) -1 nên m = -1 KTM

\(\Rightarrow\) m = 0 thỏa mãn đk

Vậy m = 0

Chúc bn học tốt!

a: Khi m=-3 thì hệ phương trình sẽ là:

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

=>\(\left\{{}\begin{matrix}-3x+2y=1\\x+6y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=1\\3x+18y=-15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}20y=-14\\x+6y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{10}\\x=-5-6y=-5-6\cdot\dfrac{-7}{10}=\dfrac{42}{10}-5=-\dfrac{8}{10}=-\dfrac{4}{5}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

x-2y=-1

=>\(\dfrac{2m-2}{m^2+1}-\dfrac{2\cdot\left(-m^2+2m+1\right)}{2m^2+2}=1\)

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

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

=>\(m^2-3=m^2+1\)

=>-3=1(vô lý)

a: Thay m=-2 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}x-2y=-2+1=-1\\-2x+y=3\cdot\left(-2\right)-1=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y=-2\\-2x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3y=-9\\x-2y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=2y-1=2\cdot3-1=5\end{matrix}\right.\)

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

=>\(m^2\ne1\)

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

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

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

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

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

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

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

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

\(x^2-y^2=4\)

=>\(\dfrac{\left(3m+1\right)^2-\left(m-1\right)^2}{\left(m+1\right)^2}=4\)

=>\(\dfrac{9m^2+6m+1-m^2+2m+1}{\left(m+1\right)^2}=4\)

=>\(8m^2+8m+2=4\left(m+1\right)^2\)

=>\(8m^2+8m+2-4m^2-8m-4=0\)

=>\(4m^2-2=0\)

=>\(m^2=\dfrac{1}{2}\)

=>\(m=\pm\dfrac{1}{\sqrt{2}}\)

a. Bạn tự giải

b. Hệ có nghiệm khi \(m\ne2\) , khi đó hệ tương đương:

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

\(\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-2}{m-2}\\x=3-2y\end{matrix}\right.\)

Do \(x=3-2y\Rightarrow\) nếu \(y\in Z\) thì \(x\in Z\)

Mà \(y=\dfrac{-2}{m-2}\Rightarrow y\in Z\) khi \(m-2=Ư\left(2\right)\)

\(\Rightarrow m-2=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow m=\left\{0;1;3;4\right\}\)

Lời giải:
Từ PT$(1)\Rightarrow x=m+1-my$. Thay vô PT(2):

$m(m+1-my)+y=3m-1$

$\Leftrightarrow y(1-m^2)+m^2+m=3m-1$

$\Leftrightarrow y(1-m^2)=-m^2+2m-1(*)$

Để hpt có nghiệm $(x,y)$ duy nhất thì pt $(*)$ cũng phải có nghiệm $y$ duy nhất 

Điều này xảy ra khi $1-m^2\neq 0\Leftrightarrow m\neq \pm 1$
Khi đó: $y=\frac{-m^2+2m-1}{1-m^2}=\frac{-(m-1)^2}{-(m-1)(m+1)}=\frac{m-1}{m+1}$

$x=m+1-my=m+1-\frac{m(m-1)}{m+1}=\frac{3m+1}{m+1}$

Có:

$x+y=\frac{m-1}{m+1}+\frac{3m+1}{m+1}=\frac{4m}{m+1}<0$

$\Leftrightarrow -1< m< 0$

Kết hợp với đk $m\neq \pm 1$ suy ra $-1< m< 0$ thì thỏa đề.