Question

\(\left\{{}\begin{matrix}\left(x+2y+1\right)\left(x+2y+2\right)=0\\xy+y^2+3y+1=0\end{matrix}\right.\)

giải hệ trên

Answer 1

Lời giải:

Từ PT(1) \(\Rightarrow \left[\begin{matrix} x+2y+1=0\\ x+2y+2=0\end{matrix}\right.\)

Nếu \(x+2y+1=0\)

PT \((2)\Leftrightarrow y(x+2y+1)-y^2+2y+1=0\)

\(\Leftrightarrow -y^2+2y+1=0\)

\(\Leftrightarrow y=1\pm \sqrt{2}\)

+) \(y=1+\sqrt{2}\rightarrow x=-1-2y=-3-2\sqrt{2}\)

+) \(y=1-\sqrt{2}\rightarrow x=-1-2y=-3+2\sqrt{2}\)

Nếu \(x+2y+2=0\)

PT \((2)\Leftrightarrow y(x+2y+2)-y^2+y+1=0\)

\(\Leftrightarrow -y^2+y+1=0\) \(\Rightarrow y=\frac{1\pm \sqrt{5}}{2}\)

+) \(y=\frac{1+\sqrt{5}}{2}\rightarrow x=-2y-2=-3-\sqrt{5}\)

+) \(y=\frac{1-\sqrt{5}}{2}\rightarrow x=-2y-2=-3+\sqrt{5}\)

Vậy...........