1. Let us solve: \[ \frac{(x - 2)}{(x + 2)} + 6\left(\frac{x - 2}{x - 6}\right) = 1,\; x \ne -2,\,6 \]

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

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

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

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

6. Solve: \[ 6\left(\cfrac{x-2}{x-6}\right)+\cfrac{x-2}{x+2}=1 \quad [x\ne -2,6] \]

7. Let us solve: \[ \frac{x + 1}{2} + \frac{2}{x + 1} = \frac{x + 1}{3} + \frac{3}{x + 1} - \frac{5}{6}, \quad x \ne -1 \]

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

9. If \(\frac{x}{y} = \frac{a + 2}{a - 2}\), then prove that \[ \frac{x^2 - y^2}{x^2 + y^2} = \frac{4a}{a^2 + 4} \]

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

11. If \( \cot\theta = \frac{x}{y} \), then prove that \[ \frac{x\cos\theta - y\sin\theta}{x\cos\theta + y\sin\theta} = \frac{x^2 - y^2}{x^2 + y^2} \]

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

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

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

15. If \(x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\) and \(xy = 1\), prove that the simplified value of \(\frac{x^2}{y} + \frac{y^2}{x}\) is 52.

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

17. \[ \frac{x + 3}{x - 3} + \frac{x - 3}{x + 3} = 2\frac{1}{2} \] Translate: \[ \frac{x + 3}{x - 3} + \frac{x - 3}{x + 3} = 2.5 \]

18. Solve: \[ \frac{x+3}{x-3} + \frac{x-3}{x+3} = 2\frac{1}{2} \quad \text{where } x \ne 3, -3 \]

19. If \[ \frac{x^2}{by + cz} = \frac{y^2}{cz + ax} = \frac{z^2}{ax + by} = 1 \] then prove that \[ \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1 \]

20. Let us solve: \[ \frac{2}{x^2} - \frac{5}{x} + 2 = 0, \; x \ne 0 \]

21. Let us solve: \[ \frac{x}{x+1} - \frac{x+1}{x} = 2\frac{1}{12}, \; x \ne 0, -1 \]

22. If \(\cfrac{x}{y} = \cfrac{a + 2}{a - 2}\), then prove that \[ \cfrac{x^2 - y^2}{x^2 + y^2} = \cfrac{4a}{a^2 + 4} \]

23. If \[ \frac{x}{y + z} = \frac{y}{z + x} = \frac{z}{x + y} \] then prove that the value of each ratio is either \(\frac{1}{2}\) or \(-1\).

24. If \[ 1\frac{1}{2} \cdot \frac{x + \sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} + \frac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}} = 14 \] then find the value of \(x\).

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

26. If \(x=\cfrac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}\) and \(xy=1\), then the value of \(\cfrac{x^2+xy+y^2}{x^2-xy+y^2}\) is —

(a) \(\cfrac{11}{12}\) (b) \(\cfrac{12}{11}\) (c) \(\cfrac{13}{12}\) (d) \(\cfrac{14}{13}\)

27. If \(a = 3 + 2\sqrt{2}\), then what is the value of \[ \frac{a^6 + a^4 + a^2 + 1}{a^3}? \]

(a) 100 (b) 200 (c) 204 (d) 250

28. Given: \[ \theta + \phi = \frac{7\pi}{12},\quad \tan\theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3} \] \[ \phi = \frac{7\pi}{12} - \frac{\pi}{3} = \frac{7\pi}{12} - \frac{4\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4} \] \[ \tan\phi = \tan\left(\frac{\pi}{4}\right) = 1 \]

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

29. 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) \(\sqrt5\)

30. If \(\frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1\), then prove that \[\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = 1\]