$$\left\lbrace\begin{array}{l} x^3(3y-2)=-8 \\ x(y^3+2)=-6 \end{array}\right. $$

Đề bài:

\begin{equation} \label{eq:70.I} \left\lbrace\begin{array}{l} x^3(3y-2)=-8 \\ x(y^3+2)=-6 \end{array}\right. \end{equation}

Lời giải:

Xét $x=0 \Rightarrow \eqref{eq:70.I} \Leftrightarrow \left\lbrace \begin{array}{l} 0=-8 \\ 0 =-6 \end{array} \right.$ (vô lý)

Xét $x \neq 0$
\begin{align} \eqref{eq:70.I} & \Leftrightarrow \left\lbrace \begin{array}{l} 3y-2=\dfrac{-8}{x^3} \\ y^3+2=\dfrac{-6}{x} \end{array} \right. \nonumber \\ \label{eq:70.2} & \Leftrightarrow \left\lbrace \begin{array}{l} 3y+\dfrac{8}{x^3}=2 \\ -\dfrac{6}{x}-y^3=2 \end{array} \right. \end{align}

Đặt $\dfrac{2}{x}=-b$
\begin{align} \eqref{eq:70.2} & \Rightarrow \left\lbrace \begin{array}{l} 3y-b^3=2 \\ 3b-y^3=2 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} 3(y-b)+\left(y^3-b^3\right)=0 \\ 3b-y^3=2 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} (y-b)\left(y^2+yb+b^2+3\right)=0 \\ 3b-y^3=2 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} y=b \quad \left(y^2+yb+b^2+3\geqslant 3>0\right) \\ b^3-3b+2=0 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} y=b \\ (b-1)^2(b+2)=0 \end{array} \right. \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} \left\lbrace \begin{array}{l} y=1 \\ b=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} y=-2 \\ b=-2 \end{array} \right. \end{array} \right. \nonumber \\ & \Rightarrow \left[ \begin{array}{l} \left\lbrace \begin{array}{l} y=1 \\ \dfrac{2}{x}=-1 \end{array} \right. \\ \left\lbrace \begin{array}{l} y=-2 \\ \dfrac{2}{x}=2 \end{array} \right. \end{array} \right. \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} \left\lbrace \begin{array}{l} x=-2 \\ y=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} x=1 \\ y=-2 \end{array} \right. \end{array} \right. \nonumber \end{align}
$$S=\left\lbrace (-2;1); (1;-2) \right\rbrace$$

____________________________
Sau khi xét $x \neq 0$ ta còn có thể giải quyết hệ phương trình này như sau:
\begin{align} \eqref{eq:70.I} & \Leftrightarrow \left\lbrace \begin{array}{l} x^3(3y-2)=-8 \\ x^3(y^3+2)=-6x^2 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} x^3(3y-2)+x^3(y^3+2)=-8-6x^2 \\ x^3(3y-2)=-8 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} 3x^3y+x^3y^3+8+6x^2=0 \\ x^3(3y-2)=-8 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} \left(xy+2\right)\left[ 3x^2+\left( x^2y^2-2xy+4 \right)\right]=0 \\ x^3(3y-2)=-8 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} xy=-2 \quad \left(3x^2+\left( x^2y^2-2xy+4 \right)>0\right) \\ x^3(3y-2)=-8 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} y= \dfrac{-2}{x} \\ x^3\left( \dfrac{-6}{x}-2\right)=-8 \end{array} \right. \nonumber \end{align}

Ta có:
\begin{align} &\phantom{\Leftrightarrow} x^3\left( \dfrac{-6}{x}-2\right)=-8 \nonumber \\ & \Leftrightarrow -6x^2-2x^3+8 =0 \nonumber \\ & \Leftrightarrow x^3+2x^2-4=0 \nonumber \\ & \Leftrightarrow x^3+2x^2+x^2-4=0 \nonumber \\ & \Leftrightarrow (x+2)\left(x^2+x-2\right) \nonumber \\ & \Leftrightarrow (x+2)^2(x-1)=0 \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} x=-2 \stackrel{\eqref{eq:70.I}}{\Rightarrow} y=1 \\ x=1 \stackrel{\eqref{eq:70.I}}{\Rightarrow} y=-2 \end{array} \right. \nonumber \end{align}