$\begin{cases}2x^3-1=5y-5x\\x^3+y^3=1\end{cases}$
$⇔\begin{cases}2x^3-x^3-y^3=5y-5x\\x^3+y^3=1\end{cases}$
$⇔\begin{cases}x^3-y^3+5x-5y=0\\x^3+y^3=1\end{cases}$
$⇔\begin{cases}(x-y)(x^2+xy+y^2+1)=0\\x^3+y^3=1\end{cases}$
$⇔\begin{cases}x-y=0\\x^3+y^3=1\end{cases}$
$⇔\begin{cases}x=y\\x^3=\dfrac{1}{2}\end{cases}$
$⇔\begin{cases}x=y\\x=\dfrac{1}{\sqrt[3]{2}}\end{cases}$
$⇔x=y=\dfrac{1}{\sqrt[3]{2}}$
Vậy $x=y=\dfrac{1}{\sqrt[3]{2}}$
