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