$\left\{ \begin{array}{l} 2y = x^{3} +1 \\ 2x = y^{3} +1 \end{array} \right.$
$\begin{cases}2y=x^3+1 \\(x-y)(x^2-xy+y^2+2)=0 \end{cases}\Leftrightarrow \begin{cases} 2y=x^3+1 \\x=y \end{cases}\Leftrightarrow  \begin{cases}x^3-2x+1=0 \\x=y \end{cases} $
$\Leftrightarrow  \begin{cases}(x-1)(x^2+x-1)=0\\x=y \end{cases} \Leftrightarrow  \begin{cases}(x-1)(x+\frac{1+\sqrt{5}}{2})( x+\frac{1-\sqrt{5}}{2} ) \\x=y \end{cases} $ 
$\Rightarrow (x;y)=[(1;1);(- \frac{1+\sqrt{5}}{2} ;-\frac{1+\sqrt{5}}{2});( -\frac{1-\sqrt{5}}{2} ; -\frac{1-\sqrt{5}}{2} )]$ 
@nghiabg321 lấy 2 vế trừ nhau là ra bạn@viet130480 bạn chứng minh vế còn lại vô ngiệm ntn ? –  gunbk92 23-02-13 11:13 PM
sao lại có (x-y)(x^2-xy y^2 2)=0 bạn giải thích hộ mình –  mathworld1999 23-02-13 11:03 PM
$\begin{cases}2y=x^3+1 \\(x-y)(x^2-xy+y^2+2)=0 \end{cases}$
$\Leftrightarrow \begin{cases} 2y=x^3+1 \\x=y \end{cases}(1)$
$Hoặc  \begin{cases}2y=x^3 +1 \\ x^2 - xy +y^2 +2= 0\end{cases}(2)$
$(1)\Leftrightarrow  \begin{cases}x^3-2x+1=0 \\x=y \end{cases} $
$\Leftrightarrow  \begin{cases}(x-1)(x^2+x-1)=0\\x=y \end{cases}$
$\Leftrightarrow  \begin{cases}(x-1)(x+\frac{1+\sqrt{5}}{2})( x+\frac{1-\sqrt{5}}{2} ) \\x=y \end{cases} $ 
$\Rightarrow (x;y)=[(1;1);(- \frac{1+\sqrt{5}}{2} ;-\frac{1+\sqrt{5}}{2});( -\frac{1-\sqrt{5}}{2} ; -\frac{1-\sqrt{5}}{2} )]$ 
 $(2)\Leftrightarrow \begin{cases}2y=x^3 +1 \\ (x+y)^2 - 3xy +2=0 \end{cases}$
$Đặt \begin{cases}S=x+y \\ P=xy \end{cases}$; ĐK : $S^2 \geq 4P$  (Viét) (3)
$(2) ,(3) \Rightarrow \begin{cases}S^2 -3P +2=0 \\S^2\geq 4P\end{cases}\Leftrightarrow \begin{cases}P=\frac{S^2 +2}{3} \\ S^2\geq 4.\frac{S^2+2}{3}\end{cases}\Rightarrow \frac{S^2+8}{3}\leq 0\Leftrightarrow S\in\varnothing $
$\Rightarrow (2)$ vô nghiệm.

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