How many real solution does the equation $\dfrac{x^{2000}}{2001} + 2\sqrt 3 x^2 - 2\sqrt 5 x + \sqrt 3$ have?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 11
\qquad\textbf{(D)}\ 12
\qquad\textbf{(E)}\ \text{None of the preceding}
$
To determine how many real solutions the equation 

\[
\frac{x^{2000}}{2001} + 2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3} = 0
\]

has, we need to analyze the behavior of the function 

\[
f(x) = \frac{x^{2000}}{2001} + 2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3}.
\]

1. **Analyze the term \(\frac{x^{2000}}{2001}\):**
   - The term \(\frac{x^{2000}}{2001}\) is always nonnegative because \(x^{2000}\) is nonnegative for all real \(x\) (since any even power of a real number is nonnegative), and dividing by a positive constant \(2001\) keeps it nonnegative.
   - Therefore, \(\frac{x^{2000}}{2001} \geq 0\).

2. **Analyze the quadratic expression \(2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3}\):**
   - We can complete the square for the quadratic part:
     \[
     2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3} = 2\sqrt{3} \left( x^2 - \frac{\sqrt{5}}{\sqrt{3}} x \right) + \sqrt{3}.
     \]
     - Completing the square inside the parentheses:
       \[
       x^2 - \frac{\sqrt{5}}{\sqrt{3}} x = \left( x - \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 - \left( \frac{\sqrt{5}}{2\sqrt{3}} \right)^2.
       \]
     - Substituting back:
       \[
       2\sqrt{3} \left( \left( x - \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 - \left( \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 \right) + \sqrt{3} = 2\sqrt{3} \left( x - \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 - 2\sqrt{3} \left( \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 + \sqrt{3}.
       \]
     - Simplifying the constants:
       \[
       2\sqrt{3} \left( x - \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 - \frac{\sqrt{3}}{6} + \sqrt{3}.
       \]
     - Combining the constants:
       \[
       2\sqrt{3} \left( x - \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 + \frac{5\sqrt{3}}{6}.
       \]
     - Since \(\left( x - \frac{\sqrt{5}}{2\sqrt{3}} \right)^2 \geq 0\), the minimum value of the quadratic expression is \(\frac{5\sqrt{3}}{6}\), which is positive.

3. **Combine the results:**
   - The function \(f(x)\) is the sum of two nonnegative terms, \(\frac{x^{2000}}{2001}\) and \(2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3}\), both of which are nonnegative and the quadratic term is strictly positive.
   - Therefore, \(f(x) > 0\) for all real \(x\).

Since \(f(x)\) is always positive, it never equals zero. Thus, the equation 

\[
\frac{x^{2000}}{2001} + 2\sqrt{3} x^2 - 2\sqrt{5} x + \sqrt{3} = 0
\]

has no real solutions.

The final answer is \(\boxed{0}\).