Ta có:
x2+2x2y2+2y2−(x2y2+2x2)−2=0x2+2x2y2+2y2−(x2y2+2x2)−2=0
⇔x2y2−x2+2y2−2=0⇔x2y2−x2+2y2−2=0
⇔x2(y2−1)+2(y2−1)=0⇔x2(y2−1)+2(y2−1)=0
⇔(y2−1)(x2+2)=0⇔(y2−1)(x2+2)=0
Dễ thấy: x2≥0∀x⇔x2+2≥2>0x2≥0∀x⇔x2+2≥2>0 (Vô nghiệm)
⇔x⇔x tùy ý
⇔y2−1=0⇔⇔y2−1=0⇔ [y=1x=−1[y=1x=−1
Vậy x tùy ý và y=1 hoặc y=−1
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