ĐKXĐ: ...
Cộng vế với vế:
\(x+y-2\sqrt{\left(x-2\right)\left(y+1\right)}=1\)
\(\Leftrightarrow x-2+y+1-2\sqrt{\left(x-2\right)\left(y+1\right)}=0\)
- Nếu \(x-2< 0\Rightarrow y+1\le0\Rightarrow x-2+y+1-2\sqrt{\left(x-2\right)\left(y+1\right)}< 0\)
\(\Rightarrow\) pt vô nghiệm
- Nếu \(x-2\ge0\Rightarrow y+1\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\\\sqrt{y+1}=b\end{matrix}\right.\) \(\Rightarrow a^2-2ab+b^2=0\Rightarrow a=b\)
\(\Rightarrow x-2=y+1\Rightarrow x=y+3\)
\(\Rightarrow-2\left(y+3\right)+y^2+y=6\)
