Bài 5: ( 1đ ). Tìm x ,y thỏa mãn : x^2+2x^2 y^2-(x^2 y^2+2x^2)-2=0

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1. x2+2x2y2+2y2−(x2y2+2x2)−2=0x2+2x2y2+2y2−(x2y2+2×2)−2=0

   ⇔x2+2x2y2+2y2−x2y2−2x2−2=0⇔x2+2x2y2+2y2−x2y2−2×2−2=0

   ⇔x2y2−x2+2y2−2=0⇔x2y2−x2+2y2−2=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

   ⇔(x2+2)(y2−1)=0⇔(x2+2)(y2−1)=0

   ⇔(x2+2)(y−1)(y+1)=0⇔(x2+2)(y−1)(y+1)=0

   Dễ thấy: x2+2≥2>0∀xx2+2≥2>0∀x (vô nghiệm)

   ⇒[y−1=0y+1=0⇒[y−1=0y+1=0⇒[y=1y=−1

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