$\begin{array}{l}a)\quad \text{Ta có:}\\ \quad 3n^3+10n^2 - 5\\ = 3n^3 + n^2 +9n^2 + 3n - 3n - 1 - 4\\ = n^2(3n+1) + 3n(3n + 1) - (3n+1) - 4\\ = (3n+1)(n^2 +3n -1) - 4\\ \text{Do đó}\,\,3n^3+10n^2 - 5\ \vdots\ 3n+1 \Leftrightarrow 4\ \vdots\ 3n+1 \\ \Leftrightarrow 3n+1 \in Ư(4)=\{-4;-2;-1;1;2;4\}\\ \Leftrightarrow 3n \in \{-5;-3;-2;0;1;3\}\\ \Leftrightarrow n \in \left\{-\dfrac53;-1;-\dfrac23;0;\dfrac13;1\right\}\\ b)\quad \text{Ta có:}\\ \quad 10n^2 + n - 10\\ = 10n^2 -10n + 11n - 11 + 1\\ = 10n(n -1) + 11(n-1) + 1\\ = (n-1)(10n + 11) + 1\\ \text{Do đó}\,\,10n^2 + n - 10\ \vdots\ n- 1\Leftrightarrow \dfrac{1}{n-1}\in\Bbb Z\\ \Leftrightarrow n - 1 = \dfrac{1}{k} \qquad (k\in \Bbb Z^*)\\ \Leftrightarrow n = \dfrac{k+1}{k} \end{array}$
