Đặt 1/(x+1)=a
\(\sqrt{y-1}=b\)
Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}2a+b=0\\3a-2b=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+2b=0\\3a-2b=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-2a=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=-1\\\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)