Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} x^3-8x=y(y^2+2)\\ x^2=3(y^2+2)\end{matrix}\right.\)
\(\Rightarrow 3(x^3-8x)=x^2y\)
\(\Leftrightarrow x[3(x^2-8)-xy]=0\)
Vì \(x^2=3y^2+6\geq 6>0\Rightarrow xeq 0\)
Do đó suy ra \(3(x^2-8)-xy=0\Rightarrow y=\frac{3(x^2-8)}{x}\)
Thay vào pt thứ 2:
\(x^2-3\frac{9(x^2-8)^2}{x^2}=6\)
Đặt $x^2=t$ thì \(t^2-27(t-8)^2=6t\)
\(\Rightarrow t=9; t=\frac{96}{13}\)
Nếu \(t=9\Rightarrow x=\pm 3\)
\(x=3\rightarrow y=1; x=-3\rightarrow y=-1\)
Nếu \(t=\frac{96}{13}\Rightarrow x=\pm \sqrt{\frac{96}{13}}\)
\(x=\sqrt{\frac{96}{13}}\rightarrow y=-\sqrt{\frac{6}{13}}; x=-\sqrt{\frac{96}{13}}\rightarrow y=\sqrt{\frac{6}{13}}\)
