Ta có:
$\left\{\begin{array}{l}x^2+y^2=\dfrac{1}{5}\\4x^3+3x-57=-y(3x+1)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^2+y^2=\dfrac{1}{5}\\25(x^2+y^2)+200x^2+150x-114=5-50y(3x+1)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^2+y^2=\dfrac{1}{2}\\25(3x+y)^2+50(3x+y)-119=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x^2+y^2=\dfrac{1}{5}\\3x+y=-\dfrac{17}{5}\end{array}\right.\\\left\{\begin{array}{l}x^2+y^2=\dfrac{1}{5}\\3x+y=\dfrac{7}{5}\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=\dfrac{2}{5}\\y=\dfrac{1}{5}\end{array}\right.\\\left\{\begin{array}{l}x=\dfrac{11}{25}\\y=\dfrac{2}{25}\end{array}\right.\end{array}\right.$