Simplify radical expressions by rationalizing the denominator.

Let's simplify each radical expression by rationalizing the denominator. Rationalizing means eliminating radicals from the denominator, usually by multiplying the numerator and denominator by the conjugate of the denominator (when it's a binomial with radicals) or by the radical itself (when it's a single radical).

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13. $\frac{7}{2 - \sqrt{6}}$

Multiply numerator and denominator by the conjugate $2 + \sqrt{6}$:

$$
\frac{7}{2 - \sqrt{6}} \cdot \frac{2 + \sqrt{6}}{2 + \sqrt{6}} = \frac{7(2 + \sqrt{6})}{(2)^2 - (\sqrt{6})^2} = \frac{7(2 + \sqrt{6})}{4 - 6} = \frac{7(2 + \sqrt{6})}{-2}
$$

$$
= -\frac{7(2 + \sqrt{6})}{2} = -\frac{14 + 7\sqrt{6}}{2}
$$

✔ Answer: $-\frac{14 + 7\sqrt{6}}{2}$

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14. $\frac{5 - \sqrt{3}}{3 + \sqrt{5}}$

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

$$
\frac{5 - \sqrt{3}}{3 + \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{(5 - \sqrt{3})(3 - \sqrt{5})}{(3)^2 - (\sqrt{5})^2} = \frac{(5 - \sqrt{3})(3 - \sqrt{5})}{9 - 5} = \frac{(5 - \sqrt{3})(3 - \sqrt{5})}{4}
$$

Now expand the numerator:

$$
(5)(3) = 15,\quad 5(-\sqrt{5}) = -5\sqrt{5},\quad (-\sqrt{3})(3) = -3\sqrt{3},\quad (-\sqrt{3})(-\sqrt{5}) = \sqrt{15}
$$

So numerator: $15 - 5\sqrt{5} - 3\sqrt{3} + \sqrt{15}$

$$
\Rightarrow \frac{15 - 5\sqrt{5} - 3\sqrt{3} + \sqrt{15}}{4}
$$

✔ Answer: $\frac{15 - 5\sqrt{5} - 3\sqrt{3} + \sqrt{15}}{4}$

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15. $\frac{3\sqrt{2}}{\sqrt{3}}$

Multiply numerator and denominator by $\sqrt{3}$ to rationalize:

$$
\frac{3\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{2} \cdot \sqrt{3}}{3} = \frac{3\sqrt{6}}{3} = \sqrt{6}
$$

✔ Answer: $\sqrt{6}$

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16. $\frac{5}{2 - \sqrt{5}}$

Multiply numerator and denominator by the conjugate $2 + \sqrt{5}$:

$$
\frac{5}{2 - \sqrt{5}} \cdot \frac{2 + \sqrt{5}}{2 + \sqrt{5}} = \frac{5(2 + \sqrt{5})}{(2)^2 - (\sqrt{5})^2} = \frac{5(2 + \sqrt{5})}{4 - 5} = \frac{5(2 + \sqrt{5})}{-1} = -5(2 + \sqrt{5})
$$

$$
= -10 - 5\sqrt{5}
$$

✔ Answer: $-10 - 5\sqrt{5}$

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17. $\frac{\sqrt{5}}{6 - \sqrt{5}}$

Multiply numerator and denominator by the conjugate $6 + \sqrt{5}$:

$$
\frac{\sqrt{5}}{6 - \sqrt{5}} \cdot \frac{6 + \sqrt{5}}{6 + \sqrt{5}} = \frac{\sqrt{5}(6 + \sqrt{5})}{(6)^2 - (\sqrt{5})^2} = \frac{6\sqrt{5} + 5}{36 - 5} = \frac{6\sqrt{5} + 5}{31}
$$

✔ Answer: $\frac{5 + 6\sqrt{5}}{31}$

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18. $\frac{6\sqrt{3}}{\sqrt{2}}$

Multiply numerator and denominator by $\sqrt{2}$:

$$
\frac{6\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{6}}{2} = 3\sqrt{6}
$$

✔ Answer: $3\sqrt{6}$

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19. $\frac{8 - \sqrt{2}}{2 - \sqrt{3}}$

Multiply numerator and denominator by the conjugate $2 + \sqrt{3}$:

$$
\frac{8 - \sqrt{2}}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{(8 - \sqrt{2})(2 + \sqrt{3})}{(2)^2 - (\sqrt{3})^2} = \frac{(8 - \sqrt{2})(2 + \sqrt{3})}{4 - 3} = (8 - \sqrt{2})(2 + \sqrt{3})
$$

Now expand:

$$
8 \cdot 2 = 16,\quad 8 \cdot \sqrt{3} = 8\sqrt{3},\quad -\sqrt{2} \cdot 2 = -2\sqrt{2},\quad -\sqrt{2} \cdot \sqrt{3} = -\sqrt{6}
$$

So total: $16 + 8\sqrt{3} - 2\sqrt{2} - \sqrt{6}$

✔ Answer: $16 + 8\sqrt{3} - 2\sqrt{2} - \sqrt{6}$

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20. $\frac{8}{\sqrt{5}}$

Multiply numerator and denominator by $\sqrt{5}$:

$$
\frac{8}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{8\sqrt{5}}{5}
$$

✔ Answer: $\frac{8\sqrt{5}}{5}$

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21. $\frac{\sqrt{2} - \sqrt{5}}{\sqrt{5} - \sqrt{2}}$

Notice that the numerator is the negative of the denominator:

$$
\frac{\sqrt{2} - \sqrt{5}}{\sqrt{5} - \sqrt{2}} = \frac{-(\sqrt{5} - \sqrt{2})}{\sqrt{5} - \sqrt{2}} = -1
$$

✔ Answer: $-1$

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✔ Final Answers:

| Problem | Simplified Expression |
|--------|------------------------|
| 13. | $-\frac{14 + 7\sqrt{6}}{2}$ |
| 14. | $\frac{15 - 5\sqrt{5} - 3\sqrt{3} + \sqrt{15}}{4}$ |
| 15. | $\sqrt{6}$ |
| 16. | $-10 - 5\sqrt{5}$ |
| 17. | $\frac{5 + 6\sqrt{5}}{31}$ |
| 18. | $3\sqrt{6}$ |
| 19. | $16 + 8\sqrt{3} - 2\sqrt{2} - \sqrt{6}$ |
| 20. | $\frac{8\sqrt{5}}{5}$ |
| 21. | $-1$ |

Let me know if you'd like these written in a boxed format or need further explanation!