Lời giải:
Từ PT(2) suy ra \(y^2=3(y-2x)\Rightarrow 2x=y-\frac{y^2}{3}\)
Thay vào PT(1) ta có:
\(x^2=xy+1\)
\(\Leftrightarrow 4x^2-4xy-4=0\)
\(\Leftrightarrow (y-\frac{y^2}{3})^2-2y(y-\frac{y^2}{3})-4=0\)
\(\Leftrightarrow y^2+\frac{y^4}{9}-\frac{2y^3}{3}-2y^2+\frac{2y^3}{3}-4=0\)
\(\Leftrightarrow \frac{y^4}{9}-y^2-4=0\)
\(\Leftrightarrow y^4-9y^2-36=0\)
\(\Leftrightarrow (y^2-\frac{9}{2})^2=\frac{225}{4}\)
\(\Rightarrow \left [\begin{matrix} y^2-\frac{9}{2}=\frac{15}{2}\\ y^2-\frac{9}{2}=\frac{-15}{2}\end{matrix}\right.\) \(\Leftrightarrow \left [\begin{matrix} y^2=12\\ y^2=-3(l)\end{matrix}\right.\)
\(y^2=12\Rightarrow y=\pm 2\sqrt{3}\)
Có: \(2x=y-\frac{y^2}{3}=y-4\Rightarrow x=\frac{y}{2}-2\)
Nếu \(y=2\sqrt{3}\Rightarrow x=\sqrt{3}-2\)
Nếu \(y=-2\sqrt{3}\Rightarrow x=-\sqrt{3}-2\)
Vậy \((x,y)=(\sqrt{3}-2, 2\sqrt{3}); (-\sqrt{3}-2, -2\sqrt{3})\)
