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

=>\(2m^2\ne m+1\)

=>\(2m^2-m-1\ne0\)

=>\(\left(m-1\right)\left(2m+1\right)\ne0\)

=>\(m\notin\left\{1;-\dfrac{1}{2}\right\}\)

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

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

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

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

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

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

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

Để x,y đều nguyên thì \(\left\{{}\begin{matrix}m+1⋮m-1\\2m⋮m-1\end{matrix}\right.\)

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

=>\(2⋮m-1\)

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

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

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

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

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

Pt có nghiệm duy nhất khi \(2m^2-m-1\ne0\Rightarrow m\ne\left\{1;-\dfrac{1}{2}\right\}\)

Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{2m^2-2m-1}{2m^2+3m+1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{\left(m+1\right)\left(2m+1\right)}=\dfrac{m-1}{m+1}\\y=m^2x-m^2-2m=\dfrac{-4m^2-2m}{m+1}\end{matrix}\right.\)

Để x nguyên \(\Rightarrow\dfrac{m-1}{m+1}\in Z\Rightarrow1-\dfrac{2}{m+1}\in Z\)

\(\Rightarrow\dfrac{2}{m+1}\in Z\)

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

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

Thay vào y thấy đều thỏa mãn y nguyên.

Vậy ...

1) Thay \(m=\sqrt{3}+1\) vào hệ phương trình, ta được:

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

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

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-5+2\sqrt{3}}{13}\\3x=\sqrt{3}-\dfrac{12+10\sqrt{3}}{13}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-5+2\sqrt{3}}{13}\\x=\left(\dfrac{13\sqrt{3}-12-10\sqrt{3}}{13}\right)\cdot\dfrac{1}{3}=\dfrac{3\sqrt{3}-12}{13}\cdot\dfrac{1}{3}=\dfrac{\sqrt{3}-4}{13}\end{matrix}\right.\)

Vậy: Khi \(m=\sqrt{3}+1\) thì hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{3}-4}{13}\\y=\dfrac{-5+2\sqrt{3}}{13}\end{matrix}\right.\)

a) \(\hept{\begin{cases}3x+2y=4\\2x-y=m\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+2y=4\\4x-2y=2m\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2m+4}{7}\\y=2x-m\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2m+4}{7}\\y=\frac{8-3m}{7}\end{cases}}\)

Để phương trình có nghiệm \(\left(x,y\right)\)với \(x< 1,y< 1\)thì

\(\hept{\begin{cases}\frac{2m+4}{7}< 1\\\frac{8-3m}{7}< 1\end{cases}}\Leftrightarrow\hept{\begin{cases}2m< 3\\3m>1\end{cases}}\Leftrightarrow\frac{1}{3}< m< \frac{2}{3}\).

b) Để ba đường thẳng đã cho đồng quy thì: 

\(\frac{2m+4}{7}+2.\frac{8-3m}{7}=3\Leftrightarrow m=-\frac{1}{4}\).

1.Để  đường thẳng  \(y=\left(m-1\right)x+3\) song song với đường thẳng \(y=2x+1\)

thì \(m-1=2\Rightarrow m=3\)

2. a. Với \(m=-2\Rightarrow\)\(\hept{\begin{cases}-2x-2y=3\\3x-2y=4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=-\frac{17}{10}\end{cases}}\)

b. Với \(m=0\Rightarrow\hept{\begin{cases}-2y=3\\3x=4\end{cases}\Rightarrow\hept{\begin{cases}y=-\frac{3}{2}\\x=\frac{4}{3}\end{cases}\left(l\right)}}\)

Với \(m\ne0\Rightarrow\hept{\begin{cases}m^2x-2my=3m\\6x+2my=8\end{cases}\Rightarrow\left(m^2+6\right)x=3m+8}\)

\(\Rightarrow x=\frac{3m+8}{m^2+6}\)\(\Rightarrow y=\frac{mx-3}{2}=\frac{m\left(3m+8\right)-3\left(m^2+6\right)}{2\left(m^2+6\right)}=\frac{4m-9}{m^2+6}\)

Để \(x+y=5\Rightarrow\frac{3m+8}{m^2+6}+\frac{4m-9}{m^2+6}=5\Rightarrow7m-1=5m^2+30\)

\(\Rightarrow-5m^2+7m-31=0\)

Ta thấy phương trình vô nghiệm nên không tồn tại m để \(x+y=5\)