1. Problem 30: Simplify the expression $$\frac{2}{x+1} + \frac{1}{x-2} - \frac{1}{x^2 - 1}$$. 2. First, factor the denominator of the third term: $$x^2 - 1 = (x-1)(x+1)$$. 3. Find a common denominator for all terms: it is $$(x-2)(x-1)(x+1)$$. 4. Rewrite each fraction with the common denominator: $$\frac{2}{x+1} = \frac{2(x-2)(x-1)}{(x+1)(x-2)(x-1)}$$ $$\frac{1}{x-2} = \frac{1(x-1)(x+1)}{(x-2)(x-1)(x+1)}$$ $$-\frac{1}{(x-1)(x+1)} = -\frac{(x-2)}{(x-2)(x-1)(x+1)}$$ 5. Combine the numerators: $$2(x-2)(x-1) + (x-1)(x+1) - (x-2)$$. 6. Expand each term: $$2(x^2 - 3x + 2) + (x^2 - 1) - (x - 2) = 2x^2 - 6x + 4 + x^2 - 1 - x + 2$$ 7. Simplify the numerator: $$3x^2 - 7x + 5$$ 8. So the simplified expression is: $$\frac{3x^2 - 7x + 5}{(x-2)(x-1)(x+1)}$$, which matches option (c). --- 9. Problem 31: Simplify $$\frac{1}{x-3} + \frac{1}{x+3}$$. 10. Find common denominator: $$(x-3)(x+3) = x^2 - 9$$. 11. Rewrite: $$\frac{1(x+3)}{x^2-9} + \frac{1(x-3)}{x^2-9} = \frac{x+3 + x - 3}{x^2 - 9} = \frac{2x}{x^2 - 9}$$. 12. This is option (b). --- 13. Problem 32: Simplify $$-\frac{2}{x} + \frac{1}{x-1} + \frac{1}{x+1}$$. 14. Common denominator is $$x(x-1)(x+1) = x(x^2 -1)$$. 15. Rewrite each: $$-\frac{2(x-1)(x+1)}{x(x^2-1)} + \frac{1 \cdot x (x+1)}{x(x^2-1)} + \frac{1 \cdot x(x-1)}{x(x^2-1)}$$ 16. Expand numerators: $$-2(x^2 -1) + x(x+1) + x(x-1) = -2x^2 + 2 + x^2 + x + x^2 - x$$ 17. Simplify numerator: $$(-2x^2 + x^2 + x^2) + (x - x) + 2 = 0 + 0 + 2 = 2$$ 18. So the expression is: $$\frac{2}{x(x^2 -1)}$$, matching option (c). --- 19. Problem 33: Simplify $$\frac{x^2 - 3x + 2}{x^2 + 5x + 4} \times \frac{x+1}{x-1}$$. 20. Factor all polynomials: $$x^2 -3x + 2 = (x-1)(x-2)$$ $$x^2 + 5x +4 = (x+4)(x+1)$$ 21. Substitute: $$\frac{(x-1)(x-2)}{(x+4)(x+1)} \times \frac{x+1}{x-1}$$ 22. Cancel common factors $$(x+1)$$ and $$(x-1)$$: $$\frac{(x-2)}{(x+4)}$$ 23. Matches option (c). --- 24. Problem 34: Simplify $$\frac{4x^2 -4}{x^2 + 2x + 1} \cdot \frac{x+1}{x-1}$$. 25. Factor: $$4x^2 -4 = 4(x^2 -1) = 4(x-1)(x+1)$$ $$x^2 + 2x + 1 = (x+1)^2$$ 26. Substitute: $$\frac{4(x-1)(x+1)}{(x+1)^2} \cdot \frac{x+1}{x-1}$$ 27. Cancel $$(x+1)$$ and $$(x-1)$$ terms: $$\frac{4}{x+1} \cdot 1 = \frac{4}{x+1}$$ 28. Matches option (b). --- 29. Problem 35: Simplify $$\frac{x^2-9}{x^2 + 6x +9} \cdot \frac{x+3}{x-9}$$. 30. Factor: $$x^2 - 9 = (x-3)(x+3)$$ $$x^2 + 6x +9 = (x+3)^2$$ 31. Substitute: $$\frac{(x-3)(x+3)}{(x+3)^2} \cdot \frac{x+3}{x-9} = \frac{x-3}{x+3} \cdot \frac{x+3}{x-9}$$ 32. Cancel $$(x+3)$$: $$\frac{x-3}{x-9}$$ 33. Matches option (a). --- 34. Problem 36: Simplify $$\frac{3x^2 -12}{x+2} \cdot \frac{x}{x-2}$$. 35. Factor numerator: $$3x^2 - 12 = 3(x^2 -4) = 3(x-2)(x+2)$$ 36. Substitute: $$\frac{3(x-2)(x+2)}{x+2} \cdot \frac{x}{x-2}$$ 37. Cancel $$(x+2)$$ and $$(x-2)$$: $$3 \cdot x = 3x$$ 38. Matches option (a). --- 39. Problem 37: Simplify $$\frac{2 - 3x - 2x^2}{x^2 + 3x} \div \frac{x^2 + 3x + 2}{x+3}$$. 40. Factor denominators: $$x^2 + 3x = x(x+3)$$ $$x^2 + 3x + 2 = (x+1)(x+2)$$ 41. Rewrite division as multiplication by reciprocal: $$\frac{2 - 3x - 2x^2}{x(x+3)} \times \frac{x+3}{(x+1)(x+2)}$$ 42. Cancel $$(x+3)$$: $$\frac{2 - 3x - 2x^2}{x} \times \frac{1}{(x+1)(x+2)}$$ 43. Factor numerator $2 - 3x - 2x^2$: Rewrite as: $$-2x^2 - 3x + 2$$ Try to factor: $$(2x+1)(-x+2) = -2x^2 + 3x + 2$$ which is close but sign differs. Try: $$(1 - 2x)(x + 1) = x + 1 - 2x^2 - 2x = -2x^2 - x + 1$$ no. Try grouping: No simple factorization; keep as is. 44. So expression is: $$\frac{2 - 3x - 2x^2}{x(x+1)(x+2)}$$ Rewrite numerator as $$1 - 2x - x^2$$ by rearrangement? No perfect match. Try rewriting numerator as $$1 - 2x - x^2$$: Actually original numerator is $$2 - 3x - 2x^2$$. Try factor by trial: Factorizing: roots from quadratic equation $$-2x^2 - 3x + 2 = 0$$ Multiply both sides by -1: $$2x^2 + 3x - 2 = 0$$ Roots: $$x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$$ Roots: $$x = \frac{2}{4} = \frac{1}{2}$$ or $$x = \frac{-8}{4} = -2$$ So factor: $$2x^2 + 3x - 2 = (2x -1)(x + 2)$$ Therefore original numerator: $$-(2x^2 + 3x - 2) = -(2x -1)(x + 2) = (1 - 2x)(x + 2)$$ 45. Substitute numerator: $$\frac{(1 - 2x)(x+2)}{x(x+1)(x+2)} = \frac{1 - 2x}{x(x+1)}$$ 46. Matches option (c). --- 47. Problem 38: Simplify $$\frac{x + 2}{1 + \frac{2}{x}}$$. 48. Rewrite denominator: $$1 + \frac{2}{x} = \frac{x + 2}{x}$$ 49. So expression is: $$\frac{x + 2}{\frac{x + 2}{x}} = (x + 2) \times \frac{x}{x + 2} = x$$ 50. Matches option (a).