To rationalize the expression $\frac{\sqrt{2} + \sqrt{5} - 1}{6 - 2\sqrt{10}}$, follow these steps:

1. Identify the denominator and the appropriate form for rationalization. The denominator is $6 - 2\sqrt{10}$, which suggests using its conjugate $6 + 2\sqrt{10}$.

2. Multiply numerator and denominator by the conjugate of the denominator: $\frac{\sqrt{2} + \sqrt{5} - 1}{6 - 2\sqrt{10}} \cdot \frac{6 + 2\sqrt{10}}{6 + 2\sqrt{10}}$

3. Expand the numerator and denominator: $\text{Numerator: } (\sqrt{2} + \sqrt{5} - 1)(6 + 2\sqrt{10})$ $\text{Denominator: } (6 - 2\sqrt{10})(6 + 2\sqrt{10})$

4. Simplify the denominator using the difference of squares: $(6 - 2\sqrt{10})(6 + 2\sqrt{10}) = 6^2 - (2\sqrt{10})^2 = 36 - 4 \cdot 10 = 36 - 40 = -4$

5. Expand and simplify the numerator: $(\sqrt{2} + \sqrt{5} - 1)(6 + 2\sqrt{10}) = \sqrt{2} \cdot 6 + \sqrt{2} \cdot 2\sqrt{10} + \sqrt{5} \cdot 6 + \sqrt{5} \cdot 2\sqrt{10} - 1 \cdot 6 - 1 \cdot 2\sqrt{10}$ $= 6\sqrt{2} + 2\sqrt{20} + 6\sqrt{5} + 2\sqrt{50} - 6 - 2\sqrt{10}$ $= 6\sqrt{2} + 4\sqrt{5} + 6\sqrt{5} + 10\sqrt{5} - 6 - 2\sqrt{10}$ $= 6\sqrt{2} + 20\sqrt{5} - 6 - 2\sqrt{10}$

6. Combine and simplify the entire expression: $\frac{6\sqrt{2} + 20\sqrt{5} - 6 - 2\sqrt{10}}{-4} = -\frac{6\sqrt{2}}{4} - \frac{20\sqrt{5}}{4} + \frac{6}{4} + \frac{2\sqrt{10}}{4}$ $= -\frac{3\sqrt{2}}{2} - 5\sqrt{5} + \frac{3}{2} + \frac{\sqrt{10}}{2}$

The rationalized form of the given expression is: $-\frac{3\sqrt{2}}{2} - 5\sqrt{5} + \frac{3}{2} + \frac{\sqrt{10}}{2}$

Do you want further details or have any questions?

Here are 5 related questions:

1. How do you rationalize the denominator of $\frac{7}{\sqrt{3} + 2}$?
2. What is the process to simplify $\frac{5 + \sqrt{6}}{2\sqrt{3} - 1}$?
3. How do you rationalize and simplify $\frac{1}{\sqrt{a} + \sqrt{b}}$?
4. What is the result of rationalizing the denominator of $\frac{4 + \sqrt{3}}{5 - \sqrt{2}}$?
5. Can you explain the significance of using conjugates in rationalizing denominators?

Tip: When rationalizing a denominator involving a binomial with a square root, always use the conjugate of the binomial to eliminate the square root.