\(\left\{{}\begin{matrix}x^2+y^2=2x^2y^2\\\left(x+y\right)\left(1+xy\right)=4x^2y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=2\left(xy\right)^2\\\left(x+y\right)\left(1+xy\right)=4\left(xy\right)^2\end{matrix}\right.\)
☘ Đặt \(x+y=a\text{ và }xy=b\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-2b=2b^2\\a\left(1+b\right)=4b^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{4b^2}{1+b}\right)^2-2b=2b^2\left(1\right)\\a=\dfrac{4b^2}{1+b}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow16b^4-2b\left(1+b\right)^2-2b^2\left(1+b\right)^2=0\)
☘ Trường hợp 1: b = 0
☘ Trường hợp 2: \(b\ne0\)
\(\Rightarrow8b^3-\left(1+b\right)^2-b\left(1+b\right)^2=0\)
\(\Leftrightarrow8b^3-1-2b-b^2-b-2b^2-b^3=0\)
\(\Leftrightarrow7b^3-3b^2-3b-1=0\)
\(\Leftrightarrow\left(b-1\right)\left(7b^2+4b+1\right)=0\)
\(\Rightarrow b=1\)
⚠ Tự giải tiếp nha. Mà cách này hơi dài.
