Step 1: Check the form first.
Put $x=1$ in $\dfrac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}$. The top gives $(8)(0)=0$ and the bottom gives $3+2-5=0$. So it is the $\dfrac{0}{0}$ type, and we must cancel a common factor.
Step 2: Factor the bottom.
\[ 3x^2+2x-5=(x-1)(3x+5). \]
Step 3: Break $x-1$ using the square root.
Since $x-1=(\sqrt{x}-1)(\sqrt{x}+1)$, the bottom becomes $(\sqrt{x}-1)(\sqrt{x}+1)(3x+5)$.
Step 4: Cancel $(\sqrt{x}-1)$.
The top has a $(\sqrt{x}-1)$ too, so the expression reduces to
\[ \frac{9x-1}{(\sqrt{x}+1)(3x+5)}. \]
Step 5: Now substitute $x=1$.
\[ \frac{9(1)-1}{(\sqrt1+1)(3(1)+5)}=\frac{8}{(2)(8)}=\frac{8}{16}. \]
Step 6: Simplify.
\[ \frac{8}{16}=\frac{1}{2}. \]
\[ \boxed{\dfrac{1}{2}} \]