Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}x^2+y^2-xy+4y+1=0\\y\left[7-\left(x-y\right)^2\right]=2\left(x^2+1\right)\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2+1\right)+\left(2y^2-2xy+8y\right)=0\\y\left[7-\left(x-y\right)^2\right]=2\left(x^2+1\right)\end{matrix}\right.\)

\(\Rightarrow y\left[7-\left(x-y\right)^2\right]+2y^2-2xy+8y=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\left(ktm\right)\\7-\left(x-y\right)^2+2y-2x+8=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow-\left(x-y\right)^2-2\left(x-y\right)+15=0\)

Đặt \(x-y=t\Rightarrow-t^2-2t+15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y+3\\x=y-5\end{matrix}\right.\)

Thế vào pt đầu: ....