Đặt [tex]S=x+y,P=xy(S^2\geq 4P)[/tex]
Hệ đã cho trở thành: [tex]\left\{\begin{matrix} S^2-2P=2P^2\\ S(P+1)=4P^2 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} S^2-2P-2P^2=0\\ S=\frac{4P^2}{P+1} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} (\frac{4P^2}{P+1})^2-2P-2P^2=0\\ S=\frac{4P^2}{P+1} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} 16P^4-(2P+2P^2)(P+1)^2=0\\ S=\frac{4P^2}{P+1} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2P(P-1)(7P^2+4P+1)=0\\ S=\frac{4P^2}{P+1} \end{matrix}\right.\Rightarrow P=0,S=0 hoặc P=1,S=2\Rightarrow x=y=0 hoặc x=y=1[/tex]