1. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then find the value of \[ x - \frac{1}{x} \]

2. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then find the value of \[ y^2 + \frac{1}{y^2} \]

3. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then find the value of \[ x^3 - \frac{1}{x^3} \]

4. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then find the value of \[ xy + \frac{1}{xy} \]

5. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then find the value of \[ 3x^2 - 5xy + 3y^2 \]

6. If \(x = \sqrt{3} + \sqrt{2}\) and \(y = \cfrac{1}{\sqrt{3} + \sqrt{2}}\), then find the value of \((x + y)^2 + (x - y)^2\).

7. If \(x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\) and \(y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\), then show that \[ \frac{x^2 + y^2}{x^2 - y^2} = \frac{7\sqrt{3}}{12} \]

8. If \(x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\) and \(y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\), then find the simplest value of \(\frac{x^2 - xy + y^2}{x^2 + xy + y^2}\).

9. If \(x = 2\), \(y = 3\), and \(z = 6\), then calculate the value of \[ \frac{3\sqrt{x}}{\sqrt{y} + \sqrt{z}} - \frac{4\sqrt{y}}{\sqrt{z} + \sqrt{x}} + \frac{\sqrt{z}}{\sqrt{x} + \sqrt{y}} \]

10. If \(x = 2\), \(y = 3\), and \(z = 6\), then calculate the value of \[ \frac{3\sqrt{x}}{\sqrt{y} + \sqrt{z}} - \frac{4\sqrt{y}}{\sqrt{z} + \sqrt{x}} + \frac{\sqrt{z}}{\sqrt{x} + \sqrt{y}} \]

11. If \(x = \sqrt{3} + \sqrt{2}\), then find the simplest value of \[ x^3 + \frac{1}{x^3} \]

12. If \(x = 3 + \sqrt{8}\) and \(y = 3 - \sqrt{8}\), then find the value of \(x^{-3} + y^{-3}\).

(a) 199 (b) 195 (c) 198 (d) 201

13. Sure! Here's the English translation of your math question: **If** \(x = \frac{\sqrt{3}}{2}\), then what is the value of \[ \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}? \]

(a) \(2\sqrt3\) (b) \(\cfrac{1}{\sqrt3}\) (c) \(\sqrt3\) (d) \(\sqrt5\)

14. If \(x = \cfrac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}\) and \(xy = 1\), then show that \[ \cfrac{x^2 + xy + y^2}{x^2 - xy + y^2} = \cfrac{12}{11} \]

15. If \(x = 2 + \sqrt{3}\) and \(x + y = 4\), then find the simplest value of \(xy + \frac{1}{xy}\).

16. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then prove that \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \]

17. If \(x = \cfrac{1}{2 + \sqrt{3}}\) and \(y = \cfrac{1}{2 - \sqrt{3}}\), then what is the value of \(\cfrac{1}{1 + x} + \cfrac{1}{1 + y}\)?

18. Given \(x = 3 + \sqrt{3}\) and \(y = 6\), find the value of \((x + y)^2\).

19. If \(x = \sqrt{3} + \frac{1}{\sqrt{3}}\) and \(y = \sqrt{3} - \frac{1}{\sqrt{3}}\), find the value of \(\frac{x^2}{y} + \frac{y^2}{x}\).

20. Given: \[ x = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} \quad \text{and} \quad xy = 1 \] Find the value of: \[ \frac{x^2 + 3xy + y^2}{x^2 - 3xy + y^2} \]

21. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then what is the value of \(3x^2 - 5xy + 3y^2\)?

22. If \(x = \frac{\sqrt{3}}{2}\), then find the value of \[ \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} \]

23. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then prove that \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} = \frac{z^2}{1 - c^2} \]

24. If \[ x \cot\left(\frac{\pi}{6}\right) = 2 \cos\left(\frac{\pi}{3}\right) + \frac{3}{4} \sec^2\left(\frac{\pi}{4}\right) + 4 \sin\left(\frac{\pi}{6}\right) \] then find the value of \(x\).

25. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), find the value of \(3x^2 + 5xy + 3y^2\).

26. If \(x = r\sinθ \cosϕ\), \(y = r\sinθ \sinϕ\), and \(z = r\cosθ\), then find the value of \(x^2 + y^2 + z^2\).

(a) \(r\) (b) \(5r\) (c) \(\sqrt{r}\) (d) \(r^2\)

27. If \(x = \cfrac{8ab}{a + b}\), then find the value of \[ \left(\cfrac{x + 4a}{x - 4a} + \cfrac{x + 4b}{x - 4b}\right) \]

28. If \(x = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}\) and \(xy = 1\), then show that \[ \frac{x^2 + xy + y^2}{x^2 - xy + y^2} = \frac{12}{11} \]

29. If \(x = 2 + \sqrt{3}\), then find the value of \[ x + \frac{1}{x} \]

(a) 2 (b) 2√3 (c) 4 (d) 2-√3

30. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then show that: \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \]