The expression in the image asks for the simplified form of:

$\frac{4}{\sqrt{7} + \sqrt{3}}$

To simplify this expression, we can rationalize the denominator. The idea is to eliminate the square roots in the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{7} + \sqrt{3}$ is $\sqrt{7} - \sqrt{3}$.

Steps:

1. Multiply both the numerator and denominator by $\sqrt{7} - \sqrt{3}$:

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

2. Use the difference of squares formula in the denominator:

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

3. Now, the expression becomes:

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

4. Simplify the fraction:

$\sqrt{7} - \sqrt{3}$

Thus, the simplified form of the given expression is:

$\sqrt{7} - \sqrt{3}$

Would you like more details or have any questions?

Here are some related questions to expand on this topic:

1. How do you simplify other fractions with irrational denominators?
2. What is the conjugate of a binomial involving square roots?
3. Can this method be applied to complex numbers?
4. What are the properties of square roots in algebraic expressions?
5. How do you generalize the difference of squares for higher powers?

Tip: When rationalizing denominators, always remember to use the conjugate for expressions involving square roots.