lấy (1) + 2.(2) 

sẽ có x = 2m-1 

thay vào (1) sẽ ra y = 2-m 

thay x và y vừa tìm được vào phần thỏa mãn sẽ có 2 nghiệm m = -1 hoặc m = \(\dfrac{3}{2}\) rồi thay vào tìm x và y theo 2 trường hợp 

trường hợp 1: m = -1 

thì ta tìm được x = -3 và y = 3 

trường hợp 2: m= \(\dfrac{3}{2}\)

x = 2 

y = \(\dfrac{1}{2}\)  

( mình chỉ bạn cách làm thôi nên hk có trình bày rõ bạn trình bày lại nhé)

=>2x-4y=8m-10 và 2x+y=3m

=>-5y=5m-10 và 2x+y=3m

=>y=-m+2 và 2x=3m+m-2=4m-2

=>y=-m+2 và x=2m-1

2/x-1/y=-1

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

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

=>-(2m^2-m-4m+2)=4m-5

=>2m^2-5m+2=-4m+5

=>2m^2+m-3=0

=>(2m+3)(m-1)=0

=>m=1 hoặc m=-3/2

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 \(\left\{{}\begin{matrix}2m+9⋮m+2\\3m+1⋮m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m+4+5⋮m+2\\3m+6-5⋮m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5⋮m+2\\-5⋮m+2\end{matrix}\right.\)

=>\(5⋮m+2\)

=>\(m+2\in\left\{1;-1;5;-5\right\}\)

=>\(m\in\left\{-1;-3;3;-7\right\}\)

a: Để hệ có nghiệm duy nhất thì \(\frac{m}{1}<>\frac{-1}{4\left(m+1\right)}\)

=>\(4m\left(m+1\right)<>-1\)

=>\(4m^2+4m+1<>0\)

=>\(\left(2m+1\right)^2<>0\)

=>2m+1<>0

=>m<>-1/2

\(\begin{cases}mx-y=1\\ x+4\left(m+1\right)y=4m\end{cases}\Rightarrow\begin{cases}y=mx-1\\ x+4\left(m+1\right)\left(mx-1\right)=4m\end{cases}\)

=>\(\begin{cases}y=mx-1\\ x+\left(4m+4\right)\left(mx-1\right)=4m\end{cases}\Rightarrow\begin{cases}y=mx-1\\ x+4m^2x-4m+4mx-4=4m\end{cases}\)

=>\(\begin{cases}y=mx-1\\ x\left(4m^2+4m+1\right)=4m+4m+4\end{cases}\Rightarrow\begin{cases}y=mx-1\\ x\left(2m+1\right)^2=8m+4=4\left(2m+1\right)\end{cases}\)

=>\(\begin{cases}x=\frac{4}{2m+1}\\ y=mx-1=\frac{4m}{2m+1}-1=\frac{4m-2m-1}{2m+1}=\frac{2m-1}{2m+1}\end{cases}\)

Để x,y nguyên thì 4⋮2m+1 và 2m-1⋮2m+1

=>4⋮2m+1 và 2m+1-2⋮2m+1

=>4⋮2m+1 và -2⋮2m+1

=>2m+1∈Ư(2)

mà 2m+1 lẻ

nên 2m+1∈{1;-1}

=>2m∈{0;-2}

=>m∈{0;-1}

b: Để hệ có nghiệm duy nhất thì \(\frac{m+1}{2}<>\frac{3m+1}{m+2}\)

=>\(\left(m+1\right)\left(m+2\right)<>2\left(3m+1\right)\)

=>\(m^2+3m+2-6m-2<>0\)

=>\(m^2-3m<>0\)

=>m(m-3)<>0

=>m∉{0;3}

\(\begin{cases}\left(m+1\right)x+\left(3m+1\right)y=2-m\\ 2x+\left(m+2\right)y=4\end{cases}\Rightarrow\begin{cases}\left(2m+2\right)x+\left(6m+2\right)y=4-2m\\ \left(2m+2\right)x+\left(m+2\right)\left(m+1\right)y=4\left(m+1\right)\end{cases}\)

=>\(\begin{cases}x\left(2m+2\right)+y\left(m^2+3m+2\right)-\left(2m+2\right)x-\left(6m+2\right)y=4\left(m+1\right)-4+2m\\ 2x+\left(m+2\right)y=4\end{cases}\)

=>\(\begin{cases}y\left(m^2-3m\right)=6m\\ 2x+\left(m+2\right)y=4\end{cases}\Rightarrow\begin{cases}y=\frac{6}{m-3}\\ 2x=4-\left(m+2\right)\cdot y=4-\frac{6\left(m+2\right)}{m-3}=\frac{4m-12-6m-12}{m-3}=\frac{-2m-24}{m-3}\end{cases}\)

=>\(\begin{cases}y=\frac{6}{m-3}\\ x=\frac{-m-12}{m-3}\end{cases}\)

Để x,y nguyên thì 6⋮m-3 và -m-12⋮m-3

=>6⋮m-3 và -m+3-15⋮m-3

=>6⋮m-3 và -15⋮m-3

=>m-3∈ƯC(6;-15)

=>m-3∈Ư(3)

=>m-3∈{1;-1;3;-3}

=>m∈{4;2;6;0}

mà m∉{0;3}

nên m∈{2;4;6}

Vì \(\dfrac{1}{2}\ne\dfrac{-2}{3}\)

nên hệ luôn có nghiệm duy nhất

a: \(\left\{{}\begin{matrix}x-2y=-3m-4\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y=-6m-8\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y-2x-3y=-6m-8-8m+1\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=-14m-7\\2x=8m-1-3y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2m+1\\2x=8m-1-6m-3=2m-4\end{matrix}\right.\)

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

Đặt \(A=y^2+3x-1\)

\(=\left(2m+1\right)^2+3\left(m-2\right)-1\)

\(=4m^2+4m+1+3m-6-1\)

\(=4m^2+7m-6\)

\(=4\left(m^2+\dfrac{7}{4}m-\dfrac{3}{2}\right)\)

\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{8}+\dfrac{49}{64}-\dfrac{145}{64}\right)\)

\(=4\left(m+\dfrac{7}{8}\right)^2-\dfrac{145}{16}>=-\dfrac{145}{16}\)
Dấu '=' xảy ra khi m=-7/8

b: Đặt B=x^2-y^2

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

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

\(=-3m^2-8m+3\)

\(=-3\left(m^2+\dfrac{8}{3}m-1\right)\)

\(=-3\left(m^2+2\cdot m\cdot\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{25}{9}\right)\)

\(=-3\left(m+\dfrac{4}{3}\right)^2+\dfrac{25}{3}< =\dfrac{25}{3}\)

Dấu '=' xảy ra khi m=-4/3

Hệ đã cho vô nghiệm khi:

\(\dfrac{1}{2}=\dfrac{m}{3m-1}\ne\dfrac{6}{3}\)

\(\Rightarrow3m-1=2m\)

\(\Rightarrow m=1\)

Vì \(\dfrac{3}{1}\ne\dfrac{-1}{2}\)

nên hệ luôn có nghiệm duy nhất

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

=>\(\left\{{}\begin{matrix}3x-y=2m-1\\3x+6y=9m+6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=2m-1-9m-6=-7m-7\\x+2y=3m+2\end{matrix}\right.\)

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

\(y-\sqrt{x}=1\)

=>\(m+1-\sqrt{m}=1\)

=>\(m-\sqrt{m}=0\)

=>\(\sqrt{m}\left(\sqrt{m}-1\right)=0\)

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

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

=>\(m-1\ne2m\)

=>\(m\ne-1\)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)

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

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

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

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

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\left(-m-1\right)=-m^2-2m-1=-\left(m+1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-m-5\\x\cdot\left(-1\right)\cdot\left(m+1\right)=-\left(m+1\right)^2\end{matrix}\right.\)

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

\(x^2-y^2=24\)

=>\(\left(m+1\right)^2-\left(m-3\right)^2=24\)

=>\(m^2+2m+1-m^2+6m-9=24\)

=>8m-8=24

=>m=4(nhận)