\(y+1\Rightarrow y\) pt trở thành \(\left\{{}\begin{matrix}\sqrt{12-2x^2}=y+3\\\sqrt{2-y^2}=5-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12-2x^2=y^2+6y+9\\2-y^2=4x^2-20x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+y^2+6y-3=0\\4x^2+y^2-20x+23=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+y^2+6y-3=0\\8x^2+2y^2-40x+46=0\end{matrix}\right.\)
\(\Rightarrow10x^2-40x+3y^2+6y+43=0\)
\(\Leftrightarrow10\left(x-2\right)^2+3\left(y+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(2;-2\right)\)
