Chủ đề này có 1 trả lời

[hoangvipro1999]

$\left\{\begin{matrix} y(y^{2}+2x^{3}y+1)=10x^{3}\\y^{2}(1+4x^{6}y^{2})=20x^{6} \end{matrix}\right.$

$\left\{\begin{matrix} x^{3}+y^{3}-8x^{3}y^{3}=10\\ 4x^{3}y+\frac{3}{y}+1=y+3x^{3} \end{matrix}\right.$

$\left\{\begin{matrix} y(4x^{3}-1)=3\\ y^{3}(3x+1)=4 \end{matrix}\right.$

$\left\{\begin{matrix} x^{10}+x^{2}y^{2}(x+y^{2})=3\\ y^{10}+\frac{y^{6}}{x^{4}}+\frac{y^{2}}{x^{2}}=\frac{3}{x^{5}} \end{matrix}\right.$

[Vu Van Quy]

$\left\{\begin{matrix} y(y^{2}+2x^{3}y+1)=10x^{3}\\y^{2}(1+4x^{6}y^{2})=20x^{6} \end{matrix}\right.$

$\left\{\begin{matrix} x^{3}+y^{3}-8x^{3}y^{3}=10\\ 4x^{3}y+\frac{3}{y}+1=y+3x^{3} \end{matrix}\right.$

$\left\{\begin{matrix} y(4x^{3}-1)=3\\ y^{3}(3x+1)=4 \end{matrix}\right.$

$\left\{\begin{matrix} x^{10}+x^{2}y^{2}(x+y^{2})=3\\ y^{10}+\frac{y^{6}}{x^{4}}+\frac{y^{2}}{x^{2}}=\frac{3}{x^{5}} \end{matrix}\right.$

Hệ 3

ĐK...

$\Leftrightarrow \left\{\begin{matrix} 4x^3-1=\frac{3}{y} & \\ 3x+1=\frac{4}{y^3} & \end{matrix}\right.$

Đây là hệ đối xứng !!!