5. Let the function $f(x)=\frac{x+3}{\sqrt{x-a}}(a \in \mathbf{R})$. If $f(x)$ is an increasing function on $(1,+\infty)$, then the range of values for $a$ is $\qquad$ .
5. $(-\infty,-1]$.

Let $t=\sqrt{x-a}$, then the original function becomes
$$
f(t)=\frac{t^{2}+3+a}{t}=t+\frac{3+a}{t} \text {. }
$$

Since $f(x)$ is an increasing function on $(1,+\infty)$, $f(t)$ is an increasing function on $(\sqrt{1-a},+\infty)$.
When $3+a \leqslant 0$, i.e., $a \leqslant-3$, it clearly meets the requirement;
When $3+a>0$, since $f(t)$ is an increasing function on $(\sqrt{3+a},+\infty)$, thus, $\sqrt{3+a} \leqslant \sqrt{1-a}$, i.e., $-3<a \leqslant-1$. In summary, the range of $a$ is $(-\infty,-1]$.