Áp dụng tcdtsbn:
\(4x=-11y\Rightarrow\dfrac{x}{-11}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{121}=\dfrac{y^2}{16}=\dfrac{x^2-3y^2}{121-48}=\dfrac{803}{73}=11\\ \Rightarrow\left\{{}\begin{matrix}x^2=1331\\y^2=176\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm11\sqrt{11}\\y=\pm4\sqrt{11}\end{matrix}\right.\)
Đặt \(\dfrac{x}{-11}=\dfrac{y}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-11k\\y=4k\end{matrix}\right.\)
Ta có: \(x^2-3y^2=803\)
\(\Leftrightarrow121k^2-3\cdot16k^2=803\)
\(\Leftrightarrow73k^2=803\)
\(\Leftrightarrow k^2=11\)
Trường hợp 1: \(k=-\sqrt{11}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-11k=11\sqrt{11}\\y=4k=-4\sqrt{11}\end{matrix}\right.\)
Trường hợp 2: \(k=+\sqrt{11}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-11k=-11\sqrt{11}\\y=4k=4\sqrt{11}\end{matrix}\right.\)
