$\rm Đáp \ án:\\ (x;y)\in{(-4;3);(0;-1);(2;1);(-2;5)}\\\text{Giải thích các bước giải:}\\xy-2x+y+1=0\\⇒xy-2x+y+(3-2)=0\\⇒xy-2x+y+3-2=0\\⇒x(y-2)+y-2=-3\\⇒x(y-2)+(y-2)=-3\\⇒(y-2)(x+1)=-3\\ Ta \ có:\\-3=1.(-3)=(-3).1=(-1).3=3.(-1)\\\text{Ta có bảng sau:}\\\begin{array}{|c|c|}\hline y-2&1&-3&-1&3\\\hline x+1&-3&1&3&-1\\\hline y&3&-1&1&5\\\hline x&-4&0&2&-2\\\hline\end{array}\\ Vậy \ (x;y)\in{(-4;3);(0;-1);(2;1);(-2;5)}$
