\(\left\{{}\begin{matrix}x^2+y=6\left(1\right)\\y^2+x=6\left(2\right)\end{matrix}\right.\)
Ta trừ (1) và (2) \(\Rightarrow x^2-y^2+y-x=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=1-y\end{matrix}\right.\)
Với x = y \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2+x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left[{}\begin{matrix}x=y=2\\x=y=-3\end{matrix}\right.\end{matrix}\right.\)
Với x = 1 - y \(\Leftrightarrow\left\{{}\begin{matrix}x=1-y\\y^2-y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\frac{1-\sqrt{21}}{2}\\y=\frac{1+\sqrt{21}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{1+\sqrt{21}}{2}\\y=\frac{1-\sqrt{21}}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt có 4 nghiệm
