a) M=$\frac{9x+5}{3x-1}$ = 3+ $\frac{8}{3x-1}$
⇒ Để M nguyên mà x nguyên ⇔ 8 chia hết cho 3x-1
⇒ 3x-1 ∈ {1,-1,2,-2,4,-4,8,-8}
⇒ x ∈ {0,1,-1,3} (vì x nguyên)
b) A(1)=B(-2)⇒12+2a+$a^{2}$ = 8+2║2a+3║+$a^{2}$
⇒2a+4 = 2║2a+3║
⇒\(\left[ \begin{array}{l}2a+3=a+2\\-2a-3=a+2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}a=-1\\a=\frac{5}{3} \end{array} \right.\)
