\(A=\dfrac{x-5}{x-11}=1+\dfrac{6}{x-11}\)
Để `A` có giá trị nguyên thì \(\dfrac{6}{x-11} \in Z\)
\(=>x-11 \in Ư_{6}\)
Mà \(Ư_{6}=\){\(\pm 1 ;\pm 2;\pm 3;\pm 6\)}
x-11 | -1 | 1 | -2 | 2 | -3 | 3 | -6 | 6 |
x | 10 | 12 | 9 | 13 | 8 | 14 | 5 | 17 |
\(=>x \in \){`10;12;9;13;8;14;5;17`}