$\left \{ {{x(x²-2)+x²y + 4 =2(x²+y)(1)} \atop {x²-y+2=0}} \right.$
Ta xét PT (1)
$(1) ⇔ x(x² - 2) + x²y + 4 - 2x² - 2y = 0$
$⇔ x(x² - 2) + x²y - 2y - 2x² + 4 = 0$
$⇔ x(x² - 2) + y(x² - 2) - 2(x² - 2) = 0$
$⇔ (x² - 2)(x + y - 2) = 0$
$⇔\left[ \begin{array}{l}x²-2=0\\x+y-2=0\end{array} \right.$
TH1: $x²-2=0$ , ta có HPT:
$\left \{ {{x²-2=0} \atop {x²-y+2=0}} \right.$
$⇔ \left \{ {{x²-2=0} \atop {y-4=0}} \right.$
$⇔ \left \{ {{x²=2} \atop {y=4}} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{x=\sqrt{2}} \atop {y=4}} \right.\\\left \{ {{x=-\sqrt{2}} \atop {y=4}} \right.\end{array} \right.$
TH2: $x + y - 2 = 0$, ta có HPT:
$\left \{ {{x + y - 2 = 0} \atop {x²-y+2=0}} \right.$
$⇔ \left \{ {{x + y - 2 = 0} \atop {x+x²=0}} \right.$
$⇔ \left \{ {{x + y - 2 = 0} \atop {x(x + 1) = 0}} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{x=0} \atop {x + y - 2 = 0}} \right.\\\left \{ {{x+1=0} \atop {x + y - 2 = 0}} \right.\end{array} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{x=0} \atop {y=2}} \right.\\\left \{ {{x=-1} \atop {y=3}} \right.\end{array} \right.$
Vậy các cặp nghiệm $(x,y)$ thỏa mãn là: $(\sqrt{2},4)$ , $(-\sqrt{2},4)$ , $(0,2)$ , $(-1,3)$
