\(Đặt:a=\sqrt{x-1};b=\dfrac{1}{y}\left(x\ge1;y\ne0\right)\\ \left\{{}\begin{matrix}2\sqrt{x-1}+\dfrac{1}{y}=4\\\sqrt{x-1}-\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=4\\a-b=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a=3\\a-b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\1-b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=\sqrt{x-1}=1\\b=\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(2;\dfrac{1}{2}\right)\)
