Đáp án:
a)
$(x+3)(x^2+1)=0$
$x+3=0$ hoặc $x^2+1=0$
Vì $x^2\geq 0 \Rightarrow x^2+1\geq1>0$
$\Rightarrow x=-3$
b)
$3x+4y-xy=15$
$(3x-12)+(4y-xy)=3$
$3(x-4)-y(x-4)=3$
$(x-4)(3-y)=3$
$\Rightarrow x-4, 3-y\in Ư(3)=\pm1,\pm3$
Ta có bảng sau:
\begin{array}{|c|c|c|c|c|} \hline x-4&-3&-1&1&3\\ \hline 3-y&-1&-3&3&1\\ \hline x&1&3&5&7\\ \hline y&4&6&0&2\\ \hline \end{array}
Do $(x,y) \in Z$ nên $(x;y)=\{(1;4),(3;6),(5;0),(7;2)\}$
