\(x^2+2x^2y^2+2y^2-\left(x^2y^2+2x^2\right)-2=0\)
\(\Leftrightarrow x^2+2x^2y^2+2y^2-x^2y^2-2x^2-2=0\)
\(\Leftrightarrow x^2y^2-x^2+2y^2-2=0\)
\(\Leftrightarrow\left(x^2y^2-x^2\right)+\left(2y^2-2\right)=0\)
\(\Leftrightarrow x^2\left(y^2-1\right)+2\left(y^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(y^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(y-1\right)\left(y+1\right)=0\)
Dễ thấy: \(x^2+2\ge2>0\forall x\) (vô nghiệm)
\(\Rightarrow\left[{}\begin{matrix}y-1=0\\y+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)
x2 + 2x2y2 + 2y2 − (x2y2+2x2) − 2=0
x2+2x2y2+2y2−x2y2−2x2−2=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
Dễ thấy: x2+2 ≥ 2 > 0∀x (không có nghiệm)
⇒y−1=0 hoặc y+1=0⇒𝑦−1=0 hoặc 𝑦+1=0⇒y=1 hoặc y=−1
