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a.
Ta có:
$\left\{\begin{array}{l}x^3+y^3=1\\x^2y+2xy^2+y^3=2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^3+y^3=1\\x^2y+2xy^2+y^3=2(x^3+y^3)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^3+y^3\\2x^3-x^2y-2xy^2+y^3=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^3+y^3=1\\(x-y)(x+y)(2x-y)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x^3+y^3=1\\x=y\end{array}\right.\\\left\{\begin{array}{l}x^3+y^3=1\\2x=y\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=\dfrac{1}{\sqrt[3]2}\\y=\dfrac{1}{\sqrt[3]2}\end{array}\right.\\\left\{\begin{array}{l}x=\dfrac{2}{\sqrt[3]9}\\y=\dfrac{2}{\sqrt[3]9}\end{array}\right.\end{array}\right.$
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