Cho số hữu tỉ \(x = \dfrac{{a - 5}}{7}.\) Với giá trị nào của $a$ thì $x$ là số nguyên dương;

\(N \subset Q\)

\(Q \subset N\)

\(Q = Z\)

\(Q \subset Z\)

\(3 \in Q\)

\(\dfrac{5}{3} \in Z\)

\(\dfrac{0}{3} \in Q\)

\(\dfrac{1}{2} \notin N\)

\(\dfrac{{11}}{{13}}\)

\( - \dfrac{{ - 1\,2}}{{15}}\)

\(\dfrac{{ - \,5}}{{ - 7}}\)

\( - \dfrac{2}{{15}}\)

\(b \in \mathbb{Z};b \ne 0\)

\(b \ne 0\)

\(b \in \mathbb{Z}\)

\(b \in \mathbb{N};b \ne 0\)

\(\dfrac{4}{{10}}\)

\( - \dfrac{6}{{15}}\)

\( - \dfrac{{ - 4}}{{10}}\)

\(\dfrac{{ - 6}}{{ - 15}}\)

\(\dfrac{{ - 15}}{{13}}; - 1;\dfrac{{ - 11}}{{13}};\dfrac{{ - 9}}{{13}};0;\dfrac{4}{{13}};\dfrac{9}{{13}}\)

\(\dfrac{{ - 15}}{{13}};\dfrac{{ - 11}}{{13}};\dfrac{{ - 9}}{{13}}; - 1;0;\dfrac{4}{{13}};\dfrac{9}{{13}}\)

\(\dfrac{{ - 9}}{{13}};\dfrac{{ - 11}}{{13}}; - 1;\dfrac{{ - 15}}{{13}};0;\dfrac{4}{{13}};\dfrac{9}{{13}}\)

\(\dfrac{9}{{13}};\dfrac{4}{{13}};0;\dfrac{{ - 9}}{{13}};\dfrac{{ - 11}}{{13}}; - 1;\dfrac{{ - 15}}{{13}}\)