Đặt x/2=y/3=k
=>x=2k; y=3k
\(x^2=-19+y^2+xy\)
\(\Leftrightarrow4k^2=-19+9k^2+6k^2\)
\(\Leftrightarrow k^2=\dfrac{19}{11}\)
Trường hợp 1: \(k=\sqrt{\dfrac{19}{11}}\)
=>\(x=2\sqrt{\dfrac{19}{11}};y=\dfrac{3\sqrt{19}}{11}\)
Trường hợp 2: \(k=-\sqrt{\dfrac{19}{11}}\)
=>\(x=-2\sqrt{\dfrac{19}{11}};y=-\dfrac{3\sqrt{19}}{11}\)