$\begin{array}{l}1)\\\quad\dfrac{n-1}{n-3}=\dfrac53\\\Leftrightarrow 3(n-1)=5(n-3)\\\Leftrightarrow 3n-3=5n-15\\\Leftrightarrow 3n-5n=-15+3\\\Leftrightarrow -2n=-12\\\Leftrightarrow n=6\\\,\\2)\\\quad xy-x+y=4\\\Leftrightarrow x(y-1)+y=4\\\Leftrightarrow x(y-1)+(y-1)=4-1\\\Leftrightarrow (x+1)(y-1)=3\\\text{mà $x,y\in\mathbb{Z}$}\\\to x+1,y-1\in Ư(3)=\{\pm1;\pm3\}\\\text{- Ta có bảng sau :}\\\begin{array}{|c|c|}\hline x+1&-3&-1&1&3\\\hline y-1&-1&-3&3&1\\\hline x&-4&-2&0&2\\\hline y&0&-2&4&2\\\hline\end{array}\\\text{- Vậy các cặp số $(x,y)$ thỏa mãn là : $(-4,0);(-2,-2);(0,4);(2,2)$} \end{array}$
