\(\left\{{}\begin{matrix}x^2+2x+1=y+xy\\2x^2+y^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=y\left(x+1\right)\\2x^2+y^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(x+1-y\right)=0\\2x^2+y^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x+1=y\end{matrix}\right.\\2x^2+y^2=9\end{matrix}\right.\)
Với x=-1 \(\Leftrightarrow y=\sqrt{7}\)
Với x+1=y\(\Leftrightarrow\)2x2+(x+1)2=9
\(\Leftrightarrow\)3x2+2x-8=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{4}{3}\rightarrow y=\dfrac{7}{3}\\x=-2\rightarrow y=-1\end{matrix}\right.\)
vậy hpt có 3 nghiệm(-1,\(\sqrt{7}\));(\(\dfrac{4}{3},\dfrac{7}{3}\));(-2,-1)
