Ta có \(\left\{{}\begin{matrix}x^2+y^2=2x^2y^2\\\left(x+y\right)\left(1+xy\right)=4x^2y^2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x+y\right)^2=2\left(xy\right)^2+2xy\\\left(x+y\right)\left(1+xy\right)=4\left(xy\right)^2\end{matrix}\right.\)(1)
Đặt a=x+y,b=xy
Vậy (1)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a^2=2b^2+2b\left(3\right)\\a\left(1+b\right)=4b^2\left(2\right)\end{matrix}\right.\)
Trong phương trình (2), nếu 1+b=0\(\Leftrightarrow b=-1\)
Vậy \(\left(2\right)\Leftrightarrow a.0=4\left(ktm\right)\)
Vậy 1+b\(\ne0\)
Vậy (2)\(\Leftrightarrow a=\dfrac{4b^2}{1+b}\)
Thay vào (3)\(\Leftrightarrow\left(\dfrac{16b^2}{1+b}\right)=2b^2+2b\Leftrightarrow16b^4=\left(2b^2+2b\right)\left(b^2+2b+1\right)\Leftrightarrow16b^4=2b^4+4b^3+2b^2+2b^3+4b^2+2b\Leftrightarrow16b^4=2b^4+6b^3+6b^2+2b\Leftrightarrow14b^4-6b^3-6b^2-2b=0\Leftrightarrow7b^4-3b^3-3b^2-b=0\Leftrightarrow b\left(7b^3-3b^2-3b-1\right)=0\Leftrightarrow b\left(7b^3-7b^2+4b^2-4b+b-1\right)=0\Leftrightarrow b\left[7b^2\left(b-1\right)+4b\left(b-1\right)+\left(b-1\right)\right]=0\Leftrightarrow b\left(b-1\right)\left(7b^2+4b+1\right)=0\)(*)
Vì 7b2+4b+1>0
(*)\(\Leftrightarrow\)\(\left[{}\begin{matrix}b=0\\b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=0\\a=2\end{matrix}\right.\)
TH1:a=0;b=0\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=0\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
TH2:a=2;b=1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy (x;y)={(0;0);(1;1)}
