Chủ đề này có 1 trả lời

[IDT]

$\left\{\begin{matrix} (x-y)(x^{2}-y^{2})+(x+y)(3xy+x-1)=-2\\ 2(x^{2}+y^{2})+3x-y=2 \end{matrix}\right.$

[Issac Newton of Ngoc Tao]

$\left\{\begin{matrix} (x-y)(x^{2}-y^{2})+(x+y)(3xy+x-1)=-2\\ 2(x^{2}+y^{2})+3x-y=2 \end{matrix}\right.$

Ta có: $HPT\Leftrightarrow \left\{\begin{matrix} (x+y)(x^{2}+y^{2}+xy+x-1)=-2 & & \\ 2(x^{2}+y^{2})+3x-y=2 & & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x+y)(2-3x+y+2xy+2x-2)=-4 & & \\ 2(x^{2}+y^{2})+3x-y=2 & & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x+y)(2xy+y-x)=-4 & & \\ 2(x^{2}+y^{2})+3x-y=2 & & \end{matrix}\right.;x+y=u;x-y=v\Rightarrow \left\{\begin{matrix} x=\frac{u+v}{2} & \\ y=\frac{u-v}{2} & \end{matrix}\right.\Rightarrow HPT:\left\{\begin{matrix} u(u^{2}-v^{2}-2v)=-8(1) & & \\ u^{2}+v^{2}+u+2v=2(2) & & \end{matrix}\right.;PT(1)+u.PT(2):2u^{3}+u^{2}-2u+8=0\Leftrightarrow u=-2\Rightarrow v=0;-2\Rightarrow (x;y)=(-2,0);(-1,-1)$.