c) \(\left\{{}\begin{matrix}3\left(x+1\right)+2y=-x\\5\left(x+y\right)=-3x+y-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+3+2y=-x\\5x+5y=-3x+y-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=-3\\8x+4y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x+4y=-6\\8x+4y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0x+0y=-1\\8x+4y=-5\end{matrix}\right.\)
Do \(0x+0y=-1\) (vô lý) nên hệ đã cho vô nghiệm
Vậy S = \(\varnothing\)
d) \(\left\{{}\begin{matrix}x+y\sqrt{5}=0\\x\sqrt{5}+3y=1-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{5}+5y=0\\x\sqrt{5}+3y=1-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{5}-1\\x\sqrt{5}+5y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}-1}{2}\\x\sqrt{5}+5\left(\dfrac{\sqrt{5}-1}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}-1}{2}\\x\sqrt{5}=\dfrac{5-5\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}-1}{2}\\x=\dfrac{\sqrt{5}-5}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\left(\dfrac{\sqrt{5}-5}{2};\dfrac{\sqrt{5}-1}{2}\right)\right\}\)