\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1+y-3\right)\left(x+1-y+3\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
TH1: x+y-2=0
=>x=-y+2
\(2x^2+y^2-6y-1=0\)
=>\(2\left(-y+2\right)^2+y^2-6y-1=0\)
=>\(2\left(y^2-4y+4\right)+y^2-6y-1=0\)
=>\(3y^2-14y+7=0\)
\(\Delta=\left(-14\right)^2-2\cdot3\cdot7=196-42=154>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}y=\dfrac{14-\sqrt{154}}{6}\\y=\dfrac{14+\sqrt{154}}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-y+2=\dfrac{-2+\sqrt{154}}{6}\\x=\dfrac{-2-\sqrt{154}}{6}\end{matrix}\right.\)
TH2: x-y+4=0
=>x=y-4
\(2x^2+y^2-6y-1=0\)
=>\(2\left(y-4\right)^2+y^2-6y-1=0\)
=>\(2\left(y^2-8y+16\right)+y^2-6y-1=0\)
=>\(3y^2-22y+31=0\)
\(\Delta=\left(-22\right)^2-4\cdot3\cdot31=112>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}y_1=\dfrac{22-\sqrt{112}}{2\cdot3}=\dfrac{11-\sqrt{28}}{3}\\y_2=\dfrac{22+\sqrt{112}}{2\cdot3}=\dfrac{11+\sqrt{28}}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=y-4=\dfrac{11-\sqrt{28}}{3}-4=\dfrac{-1-\sqrt{28}}{3}\\x=y-4=\dfrac{11+\sqrt{28}}{3}-4=\dfrac{-1+\sqrt{28}}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-y+4=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\2\cdot\left(2-y\right)^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\2\cdot\left(y-4\right)^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\3y^2-14y+7=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\3y^2-22y+31=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{7+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{7-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{11+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{11-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Vậy các cặp (x;y) thỏa mãn là: \(\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{7+2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{7-2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{11+2\sqrt{7}}{3}\right);\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{11-2\sqrt{7}}{3}\right)\)
