The expression you provided is:

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

To simplify this expression, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, $2 + \sqrt{3}$:

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

Step 1: Simplify the Denominator

The denominator becomes:

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

Step 2: Expand the Numerator

Now, expand $(4 - \sqrt{3})(2 + \sqrt{3})$:

$4 \cdot 2 + 4 \cdot \sqrt{3} - \sqrt{3} \cdot 2 - \sqrt{3} \cdot \sqrt{3} = 8 + 4\sqrt{3} - 2\sqrt{3} - 3$

Combine like terms:

$8 - 3 + (4\sqrt{3} - 2\sqrt{3}) = 5 + 2\sqrt{3}$

Final Answer

Since the denominator is 1, the expression simplifies to:

$5 + 2\sqrt{3}$

Would you like more details on any specific step?

Here are some related questions for further exploration:

1. How does multiplying by the conjugate help in simplifying radical expressions?
2. Can you apply this method to expressions with different square roots in both the numerator and denominator?
3. What is the general rule for finding the conjugate of an expression?
4. How would this process change if the denominator had a cube root instead?
5. How do rationalized denominators help in simplifying more complex algebraic expressions?

Tip: Always check for common factors before rationalizing the denominator, as it can sometimes simplify your work.