Let $\triangle ABC$ be an acute triangle, $H$ is its orthocenter. $\overrightarrow{AH},\overrightarrow{BH},\overrightarrow{CH}$ intersect $\triangle ABC$'s circumcircle at $A',B',C'$ respectively. Find the range (minimum value and the maximum upper bound) of
$$\dfrac{AH}{AA'}+\dfrac{BH}{BB'}+\dfrac{CH}{CC'}$$
1. **Given**: Let $\triangle ABC$ be an acute triangle, and $H$ be its orthocenter. The vectors $\overrightarrow{AH}, \overrightarrow{BH}, \overrightarrow{CH}$ intersect $\triangle ABC$'s circumcircle at $A', B', C'$ respectively. We need to find the range (minimum value and the maximum upper bound) of 
   \[
   \dfrac{AH}{AA'} + \dfrac{BH}{BB'} + \dfrac{CH}{CC'}
   \]

2. **Orthocenter and Altitudes**: The orthocenter $H$ is the intersection of the altitudes of $\triangle ABC$. Let $h_A, h_B, h_C$ be the lengths of the altitudes from $A, B, C$ respectively. The distance from $H$ to $BC$ is denoted as $\delta(H, BC)$.

3. **Distance Calculation**: We know that 
   \[
   \delta(H, BC) = 2R \cos(B) \cos(C)
   \]
   where $R$ is the circumradius of $\triangle ABC$.

4. **Altitude Length**: The length of the altitude from $A$ to $BC$ is given by 
   \[
   h_A = \frac{2 \cdot \text{Area of } \triangle ABC}{a} = \frac{bc \sin(A)}{a} = 2R \sin(B) \sin(C)
   \]

5. **Ratio Calculation**: The ratio $\frac{AH}{AA'}$ can be expressed as:
   \[
   \frac{AH}{AA'} = \frac{h_A - \delta(H, BC)}{h_A + \delta(H, BC)} = \frac{\cos(A)}{\cos(B-C)}
   \]

6. **Inequality**: Using the Law of Cosines on the reflection of $C$ over $AH$, we get:
   \[
   \frac{\cos(A)}{\cos(B-C)} \geq \cos(A)
   \]

7. **Expression Simplification**: We know that 
   \[
   \frac{\cos(A)}{\cos(B-C)} = \frac{b^2 + c^2 - a^2}{b^2 + c^2 - (H_A B - H_A C)^2}
   \]
   and 
   \[
   |H_A B - H_A C| = \frac{|b^2 - c^2|}{a}
   \]

8. **Further Simplification**: Thus,
   \[
   \frac{\cos(A)}{\cos(B-C)} = \frac{a^2 b^2 + a^2 c^2 - a^4}{a^2 b^2 + a^2 c^2 - b^4 - c^4 + 2b^2 c^2}
   \]

9. **Cyclic Sum**: The cyclic sum is:
   \[
   \sum_{\text{cyc}} \frac{\cos(A)}{\cos(B-C)} = \sum_{\text{cyc}} \frac{a^2 b^2 + a^2 c^2 - a^4}{a^2 b^2 + a^2 c^2 - b^4 - c^4 + 2b^2 c^2}
   \]

10. **Cauchy-Schwarz Inequality**: By the Cauchy-Schwarz inequality, we have:
    \[
    \sum_{\text{cyc}} \frac{\cos(A)}{\cos(B-C)} \geq \frac{9}{2} - 3 = \frac{3}{2}
    \]

11. **Upper Bound**: Since $\triangle ABC$ is acute, we have:
    \[
    \sum_{\text{cyc}} \frac{\cos(A)}{\cos(B-C)} < 2
    \]

12. **Conclusion**: Therefore, the range of 
    \[
    \dfrac{AH}{AA'} + \dfrac{BH}{BB'} + \dfrac{CH}{CC'}
    \]
    is 
    \[
    \frac{3}{2} \leq \dfrac{AH}{AA'} + \dfrac{BH}{BB'} + \dfrac{CH}{CC'} < 2
    \]

The final answer is \( \boxed{ \left[\frac{3}{2}, 2\right) } \)