Answer
$(-\infty, -2) \cup (-2, -1) \cup (-1, 0) \cup (0, +\infty)$
Work Step by Step
The given function is defined for all real numbers $x$ except for the ones that make the denominators equal to 0.
Note that the denominator of:
(i) $\dfrac{1}{x}$ will be zero when $x=0$;
(ii) $\dfrac{1}{x+1}$ will be zero when $x=-1$; and
(iii) $\dfrac{1}{x+2}$ will be zero when $x=-2$
Thus, the given function is defined for all real numbers $x$ except $-2, -1,$ and $0$.
In interval notation, the domain is:
$(-\infty, -2) \cup (-2, -1) \cup (-1, 0) \cup (0, +\infty)$
