a: Thay m=1 vào (I), ta được:
\(\left\{{}\begin{matrix}x+y=5\\2x-y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y+2x-y=5-2\\x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=3\\x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)
b: Để (I) có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{1}{-1}=-1\)
=>\(m\ne-2\)
\(\left\{{}\begin{matrix}mx+y=5\\2x-y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+y+2x-y=3\\2x-y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m+2\right)=3\\y=2x+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{m+2}\\y=\dfrac{6}{m+2}+2=\dfrac{6+2m+4}{m+2}=\dfrac{2m+10}{m+2}\end{matrix}\right.\)
x+y=1
=>\(\dfrac{2m+10+3}{m+2}=1\)
=>2m+13=m+2
=>m=-11(nhận)
