Let $x = \sqrt{ \dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } } + \sqrt{ \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } }$.

To simplify this expression, first let us consider each part individually: $a = \sqrt{ \dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } }$ $b = \sqrt{ \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } }$

We need to determine $x = a + b$. Let us square both $a$ and $b$ individually.

First, let's compute the product of the two fractions: $\left( \dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } \right) \cdot \left( \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \right) = 1$ since $\dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } \cdot \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } = \dfrac{ (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) }{ (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) } = \dfrac{5 - 2}{5 - 2} = 1$.

Therefore, $ab = \sqrt{ \dfrac{ (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) }{ (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) } } = 1$.

Now, consider squaring $x$: $x^2 = \left( a + b \right)^2 = a^2 + 2ab + b^2$ Since $ab = 1$, we have $2ab = 2$.

Next, we find $a^2$ and $b^2$: $a^2 = \dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} }$ $b^2 = \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

Adding these together: $a^2 + b^2 = \dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } + \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

We use the property of sum of fractions: $a^2 + b^2 = \dfrac{ (\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2 }{ (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) }$

Calculate the numerator and the denominator separately: $(\sqrt{5} - \sqrt{2})^2 = 5 - 2\sqrt{10} + 2$ $(\sqrt{5} + \sqrt{2})^2 = 5 + 2\sqrt{10} + 2$ $(\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2 = 7 - 2\sqrt{10} + 7 + 2\sqrt{10} = 14$

The denominator is: $(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 5 - 2 = 3$

Thus, $a^2 + b^2 = \dfrac{14}{3}$

Combining these results: $x^2 = a^2 + 2ab + b^2 = \dfrac{14}{3} + 2 = \dfrac{14}{3} + \dfrac{6}{3} = \dfrac{20}{3}$

So, $x = \sqrt{ \dfrac{20}{3} } = \dfrac{\sqrt{20}}{\sqrt{3}} = \dfrac{2\sqrt{5}}{\sqrt{3}} = \dfrac{2\sqrt{15}}{3}$

Therefore, the value of the expression is: $\boxed{ \dfrac{2\sqrt{15}}{3} }$

Would you like more details or have any questions? Here are some relative questions to expand your understanding:

1. How can the concept of rationalizing the denominator be applied in this problem?
2. Why is it important to verify that $ab = 1$ in this context?
3. What are some alternative methods to simplify expressions involving radicals?
4. How does squaring the terms help in solving this problem?
5. What are the properties of square roots and how are they used here?
6. How do we handle complex fractions when simplifying them?
7. Can you derive the same result using a different algebraic manipulation?
8. How can this problem be related to other mathematical concepts like symmetry in algebra?

Tip: Always verify the domain of the original expression to ensure that the operations performed are valid.