Find $$ a $$ and $$ b $$, if $$ \frac{\sqrt{5}+2}{\sqrt{5}-2}-\frac{\sqrt{5}-2}{\sqrt{5}+2}=a+b \sqrt{5} $$

Question

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To find the values of $$ a $$ and $$ b $$ such that:

$$
\frac{\sqrt{5}+2}{\sqrt{5}-2} - \frac{\sqrt{5}-2}{\sqrt{5}+2} = a + b\sqrt{5}
$$

Let's simplify each fraction by rationalizing the denominators.

1. **Rationalizing the first fraction:**

$$
\frac{\sqrt{5}+2}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{(\sqrt{5}+2)^2}{(\sqrt{5})^2 - (2)^2}
$$

$$
= \frac{5 + 4\sqrt{5} + 4}{5 - 4} = \frac{9 + 4\sqrt{5}}{1} = 9 + 4\sqrt{5}
$$

2. **Rationalizing the second fraction:**

$$
\frac{\sqrt{5}-2}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2} = \frac{(\sqrt{5}-2)^2}{(\sqrt{5})^2 - (2)^2}
$$

$$
= \frac{5 - 4\sqrt{5} + 4}{5 - 4} = \frac{9 - 4\sqrt{5}}{1} = 9 - 4\sqrt{5}
$$

3. **Subtracting the two results:**

$$
(9 + 4\sqrt{5}) - (9 - 4\sqrt{5}) = 8\sqrt{5}
$$

Comparing this with $$ a + b\sqrt{5} $$, we see that:

$$
a = 0 \quad \text{and} \quad b = 8
$$

**Answer:** $$ a = 0 $$ and $$ b = 8 $$
$$ a = 0 $$ and $$ b = 8 $$.

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