Let $x=\frac{\sqrt{6+2\sqrt5}+\sqrt{6-2\sqrt5}}{\sqrt{20}}$. The value of $$H=(1+x^5-x^7)^{{2012}^{3^{11}}}$$
is
(A)  $1$ (B)  $11$ (C) $21$ (D) $101$ (E) None of the above
1. First, we need to simplify the expression for \( x \):
   \[
   x = \frac{\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}}{\sqrt{20}}
   \]

2. We start by simplifying the terms inside the square roots. Notice that:
   \[
   \sqrt{6 + 2\sqrt{5}} = \sqrt{(\sqrt{5} + 1)^2} = \sqrt{5} + 1
   \]
   and
   \[
   \sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5} - 1)^2} = \sqrt{5} - 1
   \]

3. Substituting these back into the expression for \( x \):
   \[
   x = \frac{(\sqrt{5} + 1) + (\sqrt{5} - 1)}{\sqrt{20}}
   \]

4. Simplify the numerator:
   \[
   x = \frac{2\sqrt{5}}{\sqrt{20}}
   \]

5. Simplify the denominator:
   \[
   \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}
   \]

6. Substitute back into the expression for \( x \):
   \[
   x = \frac{2\sqrt{5}}{2\sqrt{5}} = 1
   \]

7. Now, we need to evaluate the expression \( H \):
   \[
   H = (1 + x^5 - x^7)^{2012^{3^{11}}}
   \]

8. Since \( x = 1 \):
   \[
   x^5 = 1^5 = 1 \quad \text{and} \quad x^7 = 1^7 = 1
   \]

9. Substitute these values back into the expression for \( H \):
   \[
   H = (1 + 1 - 1)^{2012^{3^{11}}} = 1^{2012^{3^{11}}}
   \]

10. Any number raised to any power is still 1:
    \[
    1^{2012^{3^{11}}} = 1
    \]

The final answer is \(\boxed{1}\).