\(\left\{{}\begin{matrix}\left|x-2\right|+2\sqrt{y+3}=9\\x+\sqrt{y+3}=-1\end{matrix}\right.\left(1\right)\)
ĐKXĐ: y>=-3
TH1: x>=2
Hệ phương trình(1) sẽ trở thành:
\(\left\{{}\begin{matrix}x-2+2\sqrt{y+3}=9\\x+\sqrt{y+3}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2\sqrt{y+3}=11\\x+\sqrt{y+3}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+3}=12\\x+\sqrt{y+3}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y+3=144\\x+12=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=144\\x=-13\left(loại\right)\end{matrix}\right.\)
=>Loại
TH2: x<2
hệ phương trình (1) sẽ trở thành \(\left\{{}\begin{matrix}-x+2+2\sqrt{y+3}=9\\x+\sqrt{y+3}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+2\sqrt{y+3}=7\\x+\sqrt{y+3}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{y+3}=6\\x+\sqrt{y+3}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y+3}=2\\x+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+3=4\\x=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\left(nhận\right)\)