15: \(\left\{{}\begin{matrix}x-y=4\\-2x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2y=8\\-2x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-2y-2x+2y=8+1\\2x-2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=9\\x-y=4\end{matrix}\right.\)
=>\(\left(x,y\right)\in\varnothing\)
16: Vì \(\dfrac{\dfrac{1}{2}}{-1}=\dfrac{1}{-2}\ne\dfrac{-2}{1}\)
nên hệ phương trình vô nghiệm
17: Vì \(\dfrac{10}{2}=\dfrac{-5}{-1}\ne\dfrac{5}{2}\)
nên hệ phương trình vô nghiệm
18: Vì \(\dfrac{1}{-2}=\dfrac{-2}{4}=\dfrac{4}{-8}\left(=-\dfrac{1}{2}\right)\)
nên hệ có vô nghiệm có dạng như sau:
\(\left\{{}\begin{matrix}x-2y=4\\-2x+4y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=4\\x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=0\\x=2y+4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\in R\\x=2y+4\end{matrix}\right.\)
19: \(\left\{{}\begin{matrix}4x-2y=\dfrac{1}{2}\\2x-y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=0,25\\2x-y=0,25\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0x=0\\y=2x-0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=2x-0,25\end{matrix}\right.\)
