The problem is as follows:

$\frac{1}{7 + 5\sqrt{3}} + \frac{1}{7 - 4\sqrt{3}}$

To simplify this expression, we should rationalize each denominator separately, then add the two fractions. Here’s a step-by-step approach:

Step 1: Rationalize each fraction

Rationalizing $\frac{1}{7 + 5\sqrt{3}}$

Multiply the numerator and denominator by the conjugate, $7 - 5\sqrt{3}$:

$\frac{1}{7 + 5\sqrt{3}} \cdot \frac{7 - 5\sqrt{3}}{7 - 5\sqrt{3}} = \frac{7 - 5\sqrt{3}}{(7 + 5\sqrt{3})(7 - 5\sqrt{3})}$

The denominator simplifies as:

$(7 + 5\sqrt{3})(7 - 5\sqrt{3}) = 49 - (5\sqrt{3})^2 = 49 - 75 = -26$

So,

$\frac{1}{7 + 5\sqrt{3}} = \frac{7 - 5\sqrt{3}}{-26} = -\frac{7 - 5\sqrt{3}}{26}$

Rationalizing $\frac{1}{7 - 4\sqrt{3}}$

Similarly, multiply the numerator and denominator by the conjugate, $7 + 4\sqrt{3}$:

$\frac{1}{7 - 4\sqrt{3}} \cdot \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{7 + 4\sqrt{3}}{(7 - 4\sqrt{3})(7 + 4\sqrt{3})}$

The denominator simplifies as:

$(7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 49 - (4\sqrt{3})^2 = 49 - 48 = 1$

So,

$\frac{1}{7 - 4\sqrt{3}} = 7 + 4\sqrt{3}$

Step 2: Add the two simplified fractions

Now we have:

$-\frac{7 - 5\sqrt{3}}{26} + (7 + 4\sqrt{3})$

To add them, convert $7 + 4\sqrt{3}$ to a fraction with a denominator of 26:

$7 + 4\sqrt{3} = \frac{26(7 + 4\sqrt{3})}{26} = \frac{182 + 104\sqrt{3}}{26}$

Now combine:

$-\frac{7 - 5\sqrt{3}}{26} + \frac{182 + 104\sqrt{3}}{26} = \frac{-7 + 5\sqrt{3} + 182 + 104\sqrt{3}}{26}$

Simplify the numerator:

$= \frac{175 + 109\sqrt{3}}{26}$

Thus, the final answer is:

$\frac{175 + 109\sqrt{3}}{26}$

Would you like more details on any of these steps?

Here are some related questions:

1. How do you find the conjugate of a binomial with a square root?
2. Why is rationalizing denominators important?
3. What are some common simplification techniques for fractions?
4. How does rationalization affect the sign of the denominator?
5. Can similar methods be used to simplify complex fractions?

Tip: When simplifying expressions with radicals, always check if the expression can be further simplified by factoring out common terms.