\(\left\{{}\begin{matrix}3x^2+y=x+3\left(1\right)\\3y^2+x=y+3\left(2\right)\end{matrix}\right.\)
Trừ vế theo vế \(\left(1\right)\) cho \(\left(2\right)\) ta có
\(3x^2+y-3y^2-x=x-y\)
\(\Leftrightarrow3x^2-3y^2+2y-2x=0\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[3\left(x+y\right)-2\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x+3y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=\dfrac{2-3y}{3}\end{matrix}\right.\)
+) Với \(x=y,\) thay vào \(\left(1\right)\) có :
\(3y^2+y=y+3\)\(\Leftrightarrow3y^2=3\Leftrightarrow y^2=1\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=1\\y=-1\Rightarrow x=-1\end{matrix}\right.\)
+) Với \(x=\dfrac{2-3y}{3}\) thay vào (1) có :
\(3\left(\dfrac{2-3y}{3}\right)^2+y=\dfrac{2-3y}{3}+3\)
\(\Leftrightarrow\left(2-3y\right)^2+3y=\left(2-3y\right)+9\)
\(\Leftrightarrow4-12y+9y^2+3y=2-3y+9\)
\(\Leftrightarrow9y^2-6x+11=0\)( vô nghiệm vì \(\Delta'=-63< 0\))
Vậy \(S=\left\{\left(1;1\right)\left(-1;-1\right)\right\}\)
