<=>\(\left\{{}\begin{matrix}x+y=2\\\left(x+y\right)\left(x^2-xy+y^2\right)=26\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x+y=2\\2\left(x^2-xy+y^2\right)=26\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x+y=2\\x^2-xy+y^2=13\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\left(x+y\right)^2=2^2\\x^2-xy+y^2=13\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x^2+2xy+y^2=4\\x^2-xy+y^2=13\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x^2+2xy+y^2=4\\\left(x^2+2xy+y^2\right)-\left(x^2-xy+y^2\right)=4-13\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x^2+2xy+y^2=4\\3xy=-9\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x^2+2xy+y^2=4\\x=-\dfrac{3}{y}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\left(x+y\right)^2=2^2\\x=-\dfrac{3}{y}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x+y=2\left(1\right)\\x=-\dfrac{3}{y}\end{matrix}\right.\)
Thay x=-3/y vào pt (1):
=>\(-\dfrac{3}{y}+y=2\)
<=>\(-3+y^2=2y\)
<=>y=3
=>x=\(-\dfrac{3}{3}=-1\)
Vậy hpt có nghiệm (x,y) là (-1;3)
