The given expression is:

$\frac{\sqrt{2}}{2\sqrt{3} + 4\sqrt{2}}$

To simplify this, let's rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, $2\sqrt{3} - 4\sqrt{2}$. The conjugate is chosen because it eliminates the radicals in the denominator upon expansion.

Step 1: Multiply numerator and denominator by the conjugate

$\frac{\sqrt{2}}{2\sqrt{3} + 4\sqrt{2}} \cdot \frac{2\sqrt{3} - 4\sqrt{2}}{2\sqrt{3} - 4\sqrt{2}}$

This becomes: $\frac{\sqrt{2}(2\sqrt{3} - 4\sqrt{2})}{(2\sqrt{3} + 4\sqrt{2})(2\sqrt{3} - 4\sqrt{2})}$

Step 2: Expand the numerator

$\sqrt{2}(2\sqrt{3} - 4\sqrt{2}) = 2\sqrt{6} - 4 \cdot \sqrt{4} = 2\sqrt{6} - 8$

So, the numerator becomes $2\sqrt{6} - 8$.

Step 3: Expand the denominator

The denominator uses the difference of squares formula: $(a + b)(a - b) = a^2 - b^2$

Here, $a = 2\sqrt{3}$ and $b = 4\sqrt{2}$. Expanding: $(2\sqrt{3})^2 - (4\sqrt{2})^2 = 4 \cdot 3 - 16 \cdot 2 = 12 - 32 = -20$

Step 4: Write the simplified expression

The fraction now becomes: $\frac{2\sqrt{6} - 8}{-20}$

Simplify the numerator and denominator: $\frac{2\sqrt{6} - 8}{-20} = \frac{-\sqrt{6} + 4}{10}$

Final Answer:

$\frac{-\sqrt{6} + 4}{10}$

Let me know if you have questions or want further clarifications!

Here are some questions related to this:

1. How does rationalizing a denominator help simplify expressions?
2. Can you use the conjugate method for complex denominators as well?
3. What is the difference of squares formula, and how is it applied here?
4. How do you handle multiple radicals in both numerator and denominator?
5. What is the significance of simplifying an expression into its simplest form?

Tip: Always remember that multiplying by the conjugate is useful when dealing with radicals in the denominator!