1: Thay m=1 vào hệ, ta được:
\(\left\{{}\begin{matrix}x+1\cdot y=1+1=2\\x\cdot1+y=2\cdot1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=2\\x+y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in R\\y=-x+2\end{matrix}\right.\)
2:
Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
\(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+m^2y=m^2+m\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(m^2-1\right)=m^2-m\\x+my=m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{m}{m+1}\\x=m+1-my=m+1-\dfrac{m^2}{m+1}=\dfrac{2m+1}{m+1}\end{matrix}\right.\)
i: x=|y|
=>\(\dfrac{2m+1}{m+1}=\left|\dfrac{m}{m+1}\right|\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}>=0\\\left(\dfrac{m}{m+1}\right)^2=\left(\dfrac{2m+1}{m+1}\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m>=-\dfrac{1}{2}\\m< -1\end{matrix}\right.\\\left(2m+1\right)^2=m^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m>=-\dfrac{1}{2}\\m< -1\end{matrix}\right.\\3m^2+4m+1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m>=-\dfrac{1}{2}\\m< -1\end{matrix}\right.\\\left(m+1\right)\left(3m+1\right)=0\end{matrix}\right.\)
=>\(m\in\varnothing\)
ii: Để x,y nguyên thì \(\left\{{}\begin{matrix}2m+1⋮m+1\\m⋮m+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+2-1⋮m+1\\m+1-1⋮m+1\end{matrix}\right.\)
=>\(-1⋮m+1\)
=>\(m+1\in\left\{1;-1\right\}\)
=>\(m\in\left\{0;-2\right\}\)
iii: 3x+4y=-5
=>\(\dfrac{3\left(2m+1\right)+4m}{m+1}=-5\)
=>\(6m+3+4m=-5m-5\)
=>10m+3=-5m-5
=>15m=-8
=>\(m=-\dfrac{8}{15}\left(nhận\right)\)
iv: x<1 và y>0
=>x-1<0 và y>0
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-m-1}{m+1}< 0\\\dfrac{m}{m+1}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{m+1}< 0\\\dfrac{m}{m+1}>0\end{matrix}\right.\)
=>\(m\in\varnothing\)
