a: Thay m=1 vào (I), ta được:
\(\left\{{}\begin{matrix}2x-y=1+1=2\\x+y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=6\\x+y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
b: Để hệ (I) có nghiệm duy nhất thì \(\dfrac{2}{1}\ne-\dfrac{m}{1}\)
=>\(m\ne-2\)
\(\left\{{}\begin{matrix}2x-my=m+1\\x+y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-my=m+1\\2x+2y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-my-2x-2y=m+1-8\\x+y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-m-2\right)=m-7\\x=4-y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-m+7}{m+2}\\x=4-y=4-\dfrac{-m+7}{m+2}=\dfrac{4m+8+m-7}{m+2}=\dfrac{5m+1}{m+2}\end{matrix}\right.\)
x-y=10
=>\(\dfrac{5m+1+m-7}{m+2}=10\)
=>10m+20=6m-6
=>4m=-26
=>m=-6,5(nhận)
