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Ta có: $\begin{cases}x^3+4y=y^3+16x\\1+y^2=5(1+x^2)\end{cases} $
$\Leftrightarrow \left\{ \begin{array}{l} x^3-y^3=4(4x-y)\\y^2-5x^2=4 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} y^2-5x^2=4\\ x^3-y^3=(y^2-5x^2)(4x-y) \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} y^2-5x^2=4\\ 21x^3-5x^2y-4xy^2=0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} y^2-5x^2=4\\ x(7x-4y)(3x+y)=0 \end{array} \right.$
*) Với $x=0$, suy ra: $y=\pm2$
*) Với $7x-4y=0$, suy ra: $-\frac{31}{16}x^2=4$, vô nghiệm.
*) Với $3x+y=0$, suy ra: $(x,y)\in\{(1,-3),(-1,3)\}$
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