\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{x+2}\left(x\ne-2\right)\\b=\sqrt{y-1}\left(y\ge1\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{1}{x+2}+2\sqrt{y-1}=5\\\dfrac{2}{x+2}+3\sqrt{y-1}=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+2b=5\\2a+3b=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2a+4b=10\\2a+3b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a+2b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=5-2.2=1\end{matrix}\right.\\ Nên:\left\{{}\begin{matrix}a=\dfrac{1}{x+2}=1\\b=\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=1\\y-1=2^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\left(TM\right)\\y=5\left(TM\right)\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-1;5\right)\)
