1. Solve: \[ \frac{x+3}{x-3} + 6\left(\frac{x-3}{x+3}\right) = 5,\quad x \ne 3,\ -3 \]

2. Solve: \[ \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} + 6\frac{6}{7} = 0 \quad \text{where } x \ne -3, 3 \]

3. Solve the equation: \[ \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} = -6\frac{6}{7}, \quad (x \ne 3, -3) \]

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

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

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

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

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

9. Let us solve: \[ \frac{12x + 17}{3x + 1} - \frac{2x + 15}{x + 7} = \frac{31}{5}, \quad x \ne -\frac{1}{3},\, -7 \]

10. Solve: \[ \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6}, \quad x \ne 1,\ 2,\ 3,\ 4 \]

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

12. If \[ x = \frac{2\sqrt{15}}{\sqrt{5} + \sqrt{3}} \] then find the value of \[ \frac{x + \sqrt{3}}{x - \sqrt{3}} + \frac{x + \sqrt{5}}{x - \sqrt{5}} \]

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

14. Let us solve: \[ (2x + 1) + \frac{3}{2x + 1} = 4,\; x \ne -\frac{1}{2} \]

15. Solve: \[ \left(\frac{x+a}{x-a}\right)^2 - 5\left(\frac{x+a}{x-a}\right) + 6 = 0,\quad x \ne a \]

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

17. Solve: \[ \frac{ax + b}{a + bx} = \frac{cx + d}{c + dx}, \quad \text{where } a \ne b,\, c \ne d,\; x \ne -\frac{a}{b},\, -\frac{c}{d} \]

18. Given: \[ x = 3 + \sqrt{3}, \quad xy = 6 \] Find the value of \((x + y)\).

19. If \(x : a = y : b = z : c\), then show that \[ \frac{x^3 + y^3 + z^3}{a^3 + b^3 + c^3} = \frac{xyz}{abc} \]