a: \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\x+\left(\sqrt{3}+1\right)y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\\left(\sqrt{3}-1\right)x+\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)y=\sqrt{3}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)x-y=\sqrt{3}\\\left(\sqrt{3}-1\right)x+2y=\sqrt{3}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3y=\sqrt{3}-\sqrt{3}+1=1\\\left(\sqrt{3}-1\right)x+2y=\sqrt{3}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\\left(\sqrt{3}-1\right)x=\sqrt{3}-1-2y=\sqrt{3}-1+\dfrac{2}{3}=\sqrt{3}-\dfrac{1}{3}=\dfrac{3\sqrt{3}-1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\dfrac{4+\sqrt{3}}{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\left(3-\sqrt{5}\right)x-3y=3+5\sqrt{5}\\4x+y=4-2\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(3-\sqrt{5}\right)x-3y=3+5\sqrt{5}\\12x+3y=12-6\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(3-\sqrt{5}+12\right)=3+5\sqrt{5}+12-6\sqrt{5}\\4x+y=4-2\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=4-2\sqrt{5}-4x=-2\sqrt{5}\end{matrix}\right.\)
