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Let's start by solving the questions from the image you've provided. Here's the breakdown:

### 6. i. If \( x = 2 - \sqrt{3} \), find the value of \( x - \frac{1}{x} \) and \( \left( x - \frac{1}{x} \right)^2 \).

1. To find \( x - \frac{1}{x} \), we first calculate \( \frac{1}{x} \).
   \[
   x = 2 - \sqrt{3} \quad \Rightarrow \quad \frac{1}{x} = \frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}
   \]
   
   Now, subtract:
   \[
   x - \frac{1}{x} = (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}
   \]

2. For \( \left( x - \frac{1}{x} \right)^2 \):
   \[
   \left( -2\sqrt{3} \right)^2 = 12
   \]

Thus, \( x - \frac{1}{x} = -2\sqrt{3} \) and \( \left( x - \frac{1}{x} \right)^2 = 12 \).

---

### 6. ii. If \( x = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} \), find the value of \( x + \frac{1}{x}, x^2 + \frac{1}{x^2}, x^3 + \frac{1}{x^3} \).

Let's first rationalize \( x \):
\[
x = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} \cdot \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{(\sqrt{5} - \sqrt{2})^2}{5 - 2} = \frac{5 - 2\sqrt{10} + 2}{3} = 1 - \frac{2\sqrt{10}}{3}
\]

Next steps will be:
1. Finding \( \frac{1}{x} \) and then adding to \( x \),
2. Calculating \( x^2 + \frac{1}{x^2} \),
3. Calculating \( x^3 + \frac{1}{x^3} \).

These require careful algebraic manipulation.

---

### 7. i. Find the value of \( x^2 + 4x + 4 \) when \( x = 2 + \sqrt{3} \).

Substitute \( x = 2 + \sqrt{3} \) into the expression:

\[
x^2 + 4x + 4 = (2 + \sqrt{3})^2 + 4(2 + \sqrt{3}) + 4
\]

First, expand \( (2 + \sqrt{3})^2 \):
\[
(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}
\]

Now, expand \( 4(2 + \sqrt{3}) \):
\[
4(2 + \sqrt{3}) = 8 + 4\sqrt{3}
\]

Now, add them:
\[
7 + 4\sqrt{3} + 8 + 4\sqrt{3} + 4 = 19 + 8\sqrt{3}
\]

Thus, \( x^2 + 4x + 4 = 19 + 8\sqrt{3} \).

---

### 7. ii. Find the value of \( 2x^2 - 3xy \) when \( x = \sqrt{2} + 3 \) and \( y = \sqrt{2} - 2 \).

Substitute these into \( 2x^2 - 3xy \):
\[
x = \sqrt{2} + 3, \quad y = \sqrt{2} - 2
\]
Calculate \( x^2 \), then multiply by 2, and also calculate \( 3xy \).

---

### 8. Find the rational numbers \( a \) and \( b \) if:

**i.** \( (8 - \sqrt{3})^2 = a + b\sqrt{3} \)

Expand the left-hand side:
\[
(8 - \sqrt{3})^2 = 64 - 16\sqrt{3} + 3 = 67 - 16\sqrt{3}
\]
Thus, \( a = 67 \) and \( b = -16 \).

I'll help with the remaining parts if you'd like. Let me know if you need more detailed steps!

### Expanding questions:
1. How do you rationalize a denominator with surds?
2. What is the general approach for solving equations involving roots like these?
3. How does one find the conjugate of a binomial with a square root?
4. Why is expanding expressions like \( (a + b)^2 \) important in these problems?
5. Can you express complex surds in terms of simpler terms?

### Tip:
When working with square roots, multiplying by the conjugate is a powerful technique to rationalize denominators or simplify expressions!

The image contains several algebraic and quadratic equations. Some involve solving expressions with square roots, rationalizing denominators, and simplifying complex surds.

This exercise focuses on solving quadratic expressions, simplifying surds, rationalizing complex denominators, and expanding binomials. It covers detailed algebraic manipulation techniques, suitable for high school students (Grades 9-12).

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