\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy+\left(x+y\right)=4\\\left(x+y+1\right)\left(5+2xy+x+y\right)=27\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}u=x+y\\v=xy\end{matrix}\right.\left(u^2\ge4v\right)\)
Khi đó hpt tt \(\left\{{}\begin{matrix}u^2-2v+u=4\\\left(u+1\right)\left(5+2v+u\right)=27\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2v=u^2+u-4\left(1\right)\\\left(u+1\right)\left(5+u^2+u-4+u\right)=27\end{matrix}\right.\)
Phương trình (1) \(\Leftrightarrow\left(u+1\right)\left(u^2+2u+1\right)=27\)
\(\Leftrightarrow u+1=\sqrt[3]{27}\) \(\Leftrightarrow u=2\)
\(\Rightarrow v=\dfrac{u^2+u-4}{2}=1\)
Khi đó\(\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Rightarrow\) x,y là nghiệm của pt: \(t^2-2t+1=0\) \(\Leftrightarrow t=1\)
\(\Rightarrow x=1;y=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2+x+y-2xy=4\\\left(x+y+1\right)\left(2xy+x+y+5\right)=27\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+u-2v=4\\\left(u+1\right)\left(2v+u+5\right)=27\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2v=u^2+u-4\\\left(u+1\right)\left(2v+u+5\right)=27\end{matrix}\right.\)
\(\Rightarrow\left(u+1\right)\left(u^2+u-4+u+5\right)=27\)
\(\Leftrightarrow\left(u+1\right)^3=27\)
\(\Leftrightarrow u+1=3\Rightarrow u=2\Rightarrow v=1\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Rightarrow x=y=1\)
