\(\left\{{}\begin{matrix}\sqrt{5}x-\left(1+\sqrt{3}\right)y=1\\\left(1-\sqrt{3}\right)x+\sqrt{5}y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{5}\left(1-\sqrt{3}\right)x-\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)y=1-\sqrt{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x+5y=\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{5}\left(1-\sqrt{3}\right)x-\left(1-3\right)y=1-\sqrt{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x+5y=\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{5}\left(1-\sqrt{3}\right)x+2y=1-\sqrt{3}\left(1\right)\\\sqrt{5}\left(1-\sqrt{3}\right)x+5y=\sqrt{5}\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\), ta được \(hpt\Leftrightarrow\left\{{}\begin{matrix}3y=\sqrt{5}+\sqrt{3}-1\\\sqrt{5}\left(1-\sqrt{3}\right)x+5y=\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}+\sqrt{3}-1}{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x+5y=\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}+\sqrt{3}-1}{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x+\dfrac{5\left(\sqrt{5}+\sqrt{3}-1\right)}{3}=\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}+\sqrt{3}-1}{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x=\sqrt{5}-\dfrac{5\left(\sqrt{5}+\sqrt{3}-1\right)}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}+\sqrt{3}-1}{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x=\dfrac{3\sqrt{5}-5\sqrt{5}-5\sqrt{3}+5}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}+\sqrt{3}-1}{3}\\\sqrt{5}\left(1-\sqrt{3}\right)x=\dfrac{5-2\sqrt{5}-5\sqrt{3}}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}+\sqrt{3}-1}{3}\\x=\dfrac{5-2\sqrt{5}-5\sqrt{3}}{3\sqrt{5}\left(1-\sqrt{3}\right)}\end{matrix}\right.\)
Tự rút gọn tiếp. :V
