ĐKXĐ: ...
\(\Leftrightarrow\left(x+1\right)^2+2\left(x+1\right).y+y^2+\sqrt{2x-y-4}=0\)
\(\Leftrightarrow\left(x+y+1\right)^2+\sqrt{2x-y-4}=0\)
Do \(\left\{{}\begin{matrix}\left(x+y+1\right)^2\ge0\\\sqrt{2x-y-4}\ge0\end{matrix}\right.\)
Nên đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x+y+1=0\\2x-y-4=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)