Question

Giaỉ hệ phương trình \(\left\{{}\begin{matrix}3x^2+y^2+4xy=8\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(3x+y\right)=8\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)

Chia vế cho vế: \(\frac{x^2+xy+2}{3x+y}=1\Leftrightarrow x^2+xy+2=3x+y\)

\(\Leftrightarrow x^2+\left(y-3\right)x-y+2=0\)

\(\Delta=\left(y-3\right)^2-4\left(-y+2\right)=\left(y+1\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{3-y+y+1}{2}=1\\x=\frac{3-y-y-1}{2}=-y+1\end{matrix}\right.\)

- Với \(x=1\Rightarrow y^2+4y-5=0\Rightarrow y=...\)

- Với \(x=-y+1\Rightarrow3\left(-y+1\right)^2+y^2+4y\left(-y+1\right)-8=0\Rightarrow y=...\)