Question

\(\left\{{}\begin{matrix}x^2-y^2+2x-4y+2=0\\xy+2x+y=4\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y+2\right)^2+5=0\\\left(x+1\right)\left(y+2\right)=6\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=a\\y+2=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2-b^2+5=0\\ab=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a^2-b^2+5=0\\b=\frac{6}{a}\end{matrix}\right.\) (do \(a=0\) ko phải nghiệm)

\(\Rightarrow a^2-\frac{36}{a^2}+5=0\Leftrightarrow a^4+5a^2-36=0\)

\(\Rightarrow\left[{}\begin{matrix}a^2=4\\a^2=-9\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=2;b=3\\a=-2;b=-3\end{matrix}\right.\)

\(\Rightarrow x=...;y=...\)