\(x\ge-3;y\ge-2\)
Từ pt đầu: \(x^2-y^2+3\left(x-y\right)=0\Leftrightarrow\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+3\right)=0\Rightarrow\left[{}\begin{matrix}x=y\\x=-3-y\end{matrix}\right.\)
TH1: \(x=y\Rightarrow x\ge-2\) thay vào pt dưới:
\(\sqrt{x+3}+\sqrt{x+2}=2\Leftrightarrow2x+5+2\sqrt{x^2+5x+6}=4\)
\(\Leftrightarrow2\sqrt{x^2+5x+6}=-2x-1\) \(\Leftrightarrow\left\{{}\begin{matrix}-2x-1\ge0\\4\left(x^2+5x+6\right)=\left(-2x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{-1}{2}\\16x=-23\end{matrix}\right.\) \(\Rightarrow x=y=\dfrac{-23}{16}\)
TH2: \(x=-3-y\Rightarrow y\le0\) thay vào pt dưới:
\(\sqrt{-y}+\sqrt{y+2}=2\) \(\Leftrightarrow-y+y+2+2\sqrt{-y^2-2y}=4\)
\(\Leftrightarrow\sqrt{-y^2-2y}=1\Leftrightarrow y^2+2y+1=0\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=-2\end{matrix}\right.\)
