Giải hệ phương trình $$\{\begin{array}{l}2\left(x^3-3x\right)=y^3+3y\\ 2\left(x^2-1\right)=5y^2+1\end{array}(x,y\in \mathbb{R}).$$

$\{\begin{array}{l}2\left(x^3-3x\right)=y^3+3y(1)\\ 2\left(x^2-1\right)=5y^2+1(2)\end{array}$

Ta có $(2)\Rightarrow y^2=\frac{2x^2-3}{5}$

suy ra $y^3+3y=y\left(y^2+3\right)=y\left(\frac{2x^2-3}{5}+3\right)=\frac{y\left(2x^2+12\right)}{5}$

do đó $(1)\iff 2\left(x^3-3x\right)=\frac{y\left(2x^2+12\right)}{5}\Rightarrow y=\frac{5\left(x^3-3x\right)}{x^2+6}$ (*)

Thay (*) vào (2) ta được: $2\left(x^2-1\right)=125\frac{{\left(x^3-3x\right)}^2}{{\left(x^2+6\right)}^2}+1$

$\iff 125{\left(x^3-3x\right)}^2-2\left(x^2-1\right){\left(x^2+6\right)}^2+{\left(x^2+6\right)}^2=0$

$\iff 125\left(x^6-6x^4+9x^2\right)-2\left(x^2-1\right)\left(x^4+12x^2+36\right)+\left(x^4+12x^2+36\right)=0$

$\iff 125\left(t^3-6t^2+9t\right)-2(t-1)\left(t^2+12t+36\right)+\left(t^2+12t+36\right)=0$ với $t=x^2\ge 0$.

$\iff 123t^3-771t^2+1089t+108=0$.

$\iff \{\begin{array}{l}t=-0,\mathrm{09...}<0(l)\\ t=4\\ t=2,\mathrm{36...}\to A\end{array}\iff \{\begin{array}{l}x=2\Rightarrow y=1\\ x=-2\Rightarrow y=-1\\ x\approx -1,5366\Rightarrow y\approx 0,5869\\ x\approx 1,5366\Rightarrow y\approx -0,5869\end{array}$.

Vậy hệ có 4 nghiệm $(x;y)$ là $(2;1)$, $(-2;-1)$, $(-1,5366;0,5869)$, $(1,5366;-0,5869)$.