Answer
False
Work Step by Step
We are given the rational function:
$$f(x)=\dfrac{7}{x(x-3)+5(x-3)}.$$
The domain of a rational function contains all real $x$ from which we exclude the zero(s) of the denominator.
Determine the zero(s) of the denominator for $f$:
$$\begin{align*}
x(x-3)+5(x-3)&=0\\
(x-3)(x+5)&=0\\
x-3=0&\text{ or }x+5=0\\
x=3&\text{ or }x=-5.
\end{align*}$$
So we must exclude the values $-5$ and $3$. The domain of $f$ is:
$$(-\infty,-5)\cup(-5,3)\cup(3,\infty).$$
Therefore the given statement is FALSE.
To make it true we can replace it by:
The domain of $f(x)=\dfrac{7}{x(x-3)+5(x-3)}$ is $(-\infty,-5)\cup(-5,3)\cup(3,\infty)$.
