1. **Problem 3:** Solve $$\frac{(x - 3)(x + 5)(x - 7)}{|x - 4|(x + 6)} \leq 0$$ 2. **Step 1:** Identify critical points where numerator or denominator is zero or undefined: $$x = 3, -5, 7, 4, -6$$ 3. **Step 2:** Note that $$|x - 4| > 0$$ for all $$x \neq 4$$, so denominator sign depends on $$x + 6$$ only. 4. **Step 3:** Analyze sign changes on intervals divided by critical points: Intervals: $$(-\infty, -6), (-6, -5), (-5, 3), (3, 4), (4, 7), (7, \infty)$$ 5. **Step 4:** Test each interval for sign of numerator and denominator: - Numerator zero at $$x=3,-5,7$$ - Denominator zero at $$x=-6,4$$ (undefined at 4) 6. **Step 5:** From sign analysis, solution is: $$x \in (-6, -5] \cup [3, 4) \cup (4, 7]$$ --- 7. **Problem 6:** Solve $$4^{-x + 0.5} - 7.2^{-x} < 4$$ 8. **Step 1:** Rewrite terms: $$4^{-x + 0.5} = 4^{0.5} \cdot 4^{-x} = 2 \cdot 4^{-x}$$ 9. **Step 2:** Let $$t = 4^{-x} > 0$$, then rewrite inequality: $$2t - 7.2^{-x} < 4$$ 10. **Step 3:** Express $$7.2^{-x}$$ in terms of $$t$$: Since $$7.2 = 4 \times 1.8$$, $$7.2^{-x} = (4 \times 1.8)^{-x} = 4^{-x} \cdot 1.8^{-x} = t \cdot 1.8^{-x}$$ 11. **Step 4:** Since $$1.8^{-x} > 0$$, inequality becomes: $$2t - t \cdot 1.8^{-x} < 4$$ 12. **Step 5:** Rearranged: $$t(2 - 1.8^{-x}) < 4$$ 13. **Step 6:** Because $$t = 4^{-x}$$ and $$1.8^{-x}$$ are positive, solve numerically or by testing values. 14. **Step 7:** Final solution: $$x \in (-2, \infty)$$ --- 15. **Problem 7:** Solve $$\frac{x^4}{(x - 2)^2} > 0$$ 16. **Step 1:** Numerator $$x^4 \geq 0$$, zero only at $$x=0$$. 17. **Step 2:** Denominator $$ (x-2)^2 > 0$$ for $$x \neq 2$$. 18. **Step 3:** Expression undefined at $$x=2$$. 19. **Step 4:** Since numerator is zero at $$x=0$$, expression equals zero there, not greater than zero. 20. **Step 5:** For $$x \neq 0, 2$$, numerator positive, denominator positive, so expression positive. 21. **Step 6:** Solution: $$x \in \mathbb{R} \setminus \{0, 2\}$$ --- 22. **Problem 8:** Solve $$\frac{6x^2 - 5x - 3}{x^2 - 2x + 6} \leq 4$$ 23. **Step 1:** Bring all terms to one side: $$\frac{6x^2 - 5x - 3}{x^2 - 2x + 6} - 4 \leq 0$$ 24. **Step 2:** Common denominator: $$\frac{6x^2 - 5x - 3 - 4(x^2 - 2x + 6)}{x^2 - 2x + 6} \leq 0$$ 25. **Step 3:** Simplify numerator: $$6x^2 - 5x - 3 - 4x^2 + 8x - 24 = 2x^2 + 3x - 27$$ 26. **Step 4:** Inequality: $$\frac{2x^2 + 3x - 27}{x^2 - 2x + 6} \leq 0$$ 27. **Step 5:** Denominator $$x^2 - 2x + 6 = (x-1)^2 + 5 > 0$$ always positive. 28. **Step 6:** Solve numerator inequality: $$2x^2 + 3x - 27 \leq 0$$ 29. **Step 7:** Find roots: $$x = \frac{-3 \pm \sqrt{9 + 216}}{4} = \frac{-3 \pm 15}{4}$$ 30. **Step 8:** Roots: $$x = 3$$ and $$x = -\frac{9}{2}$$ 31. **Step 9:** Since parabola opens upward, inequality holds between roots: $$-\frac{9}{2} \leq x \leq 3$$ --- 32. **Problem 9:** Solve $$\frac{(x + 2)(x^2 - 2x + 1)}{-4 + 3x - x^2} \geq 0$$ 33. **Step 1:** Factor numerator: $$x^2 - 2x + 1 = (x - 1)^2$$ 34. **Step 2:** Denominator: $$-4 + 3x - x^2 = -(x^2 - 3x + 4)$$ 35. **Step 3:** Quadratic $$x^2 - 3x + 4$$ has discriminant $$9 - 16 = -7 < 0$$, always positive. 36. **Step 4:** Denominator always negative: $$- (positive) = negative$$ 37. **Step 5:** Inequality becomes: $$\frac{(x + 2)(x - 1)^2}{negative} \geq 0$$ 38. **Step 6:** Multiply both sides by negative denominator (flip inequality): $$(x + 2)(x - 1)^2 \leq 0$$ 39. **Step 7:** Since $$(x - 1)^2 \geq 0$$, zero at $$x=1$$, sign depends on $$(x + 2)$$. 40. **Step 8:** Solve: $$(x + 2) \leq 0 \Rightarrow x \leq -2$$ 41. **Step 9:** Include points where expression zero: $$x = -2$$ or $$x = 1$$ 42. **Step 10:** Solution: $$x \in (-\infty, -2] \cup \{1\}$$ --- 43. **Problem 10:** Solve $$\sqrt{x + 2} \geq x$$ 44. **Step 1:** Domain: $$x + 2 \geq 0 \Rightarrow x \geq -2$$ 45. **Step 2:** Square both sides (valid since both sides non-negative in domain): $$x + 2 \geq x^2$$ 46. **Step 3:** Rearrange: $$x^2 - x - 2 \leq 0$$ 47. **Step 4:** Factor: $$(x - 2)(x + 1) \leq 0$$ 48. **Step 5:** Solution: $$-1 \leq x \leq 2$$ 49. **Step 6:** Combine with domain $$x \geq -2$$: $$x \in [-1, 2]$$ 50. **Step 7:** Check endpoints in original inequality: At $$x = -1$$, $$\sqrt{1} = 1 \geq -1$$ true. At $$x = 2$$, $$\sqrt{4} = 2 \geq 2$$ true. 51. **Step 8:** Also check $$x = -2$$: $$\sqrt{0} = 0 \geq -2$$ true. 52. **Step 9:** Since $$x \geq -2$$ and $$x \leq 2$$ from above, final solution: $$x \in [-2, 2]$$ --- 53. **Problem 11:** Solve $$\sqrt{2 + x - x^2} > x - 4$$ 54. **Step 1:** Domain: $$2 + x - x^2 \geq 0 \Rightarrow -x^2 + x + 2 \geq 0$$ 55. **Step 2:** Rewrite: $$-x^2 + x + 2 = -(x^2 - x - 2) = -(x - 2)(x + 1) \geq 0$$ 56. **Step 3:** Inequality holds where: $$(x - 2)(x + 1) \leq 0$$ 57. **Step 4:** Solution: $$-1 \leq x \leq 2$$ 58. **Step 5:** Since right side $$x - 4$$ can be negative or positive, consider cases: 59. **Step 6:** For $$x - 4 \leq 0$$ (i.e., $$x \leq 4$$), square both sides: $$2 + x - x^2 > (x - 4)^2$$ 60. **Step 7:** Expand right side: $$(x - 4)^2 = x^2 - 8x + 16$$ 61. **Step 8:** Inequality: $$2 + x - x^2 > x^2 - 8x + 16$$ 62. **Step 9:** Rearrange: $$2 + x - x^2 - x^2 + 8x - 16 > 0$$ 63. **Step 10:** Simplify: $$-2x^2 + 9x - 14 > 0$$ 64. **Step 11:** Multiply by -1 (flip inequality): $$2x^2 - 9x + 14 < 0$$ 65. **Step 12:** Discriminant: $$81 - 112 = -31 < 0$$ no real roots, quadratic always positive. 66. **Step 13:** So inequality never holds, no solution from this case. 67. **Step 14:** For $$x - 4 > 0$$ (i.e., $$x > 4$$), right side positive, left side $$\sqrt{\cdot} \geq 0$$, so inequality: $$\sqrt{2 + x - x^2} > x - 4 > 0$$ 68. **Step 15:** But domain restricts $$x \leq 2$$, so no $$x > 4$$ in domain. 69. **Step 16:** Therefore, solution is domain: $$x \in [-1, 2]$$ --- 70. **Problem 12:** Find number of integral $$x$$ satisfying: $$\sqrt{-x^2 + 10x - 16} < x - 2$$ 71. **Step 1:** Domain: $$-x^2 + 10x - 16 \geq 0$$ 72. **Step 2:** Rewrite: $$-(x^2 - 10x + 16) \geq 0 \Rightarrow x^2 - 10x + 16 \leq 0$$ 73. **Step 3:** Find roots: $$x = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2}$$ 74. **Step 4:** Roots: $$x = 2, 8$$ 75. **Step 5:** Inequality holds for: $$2 \leq x \leq 8$$ 76. **Step 6:** Inequality to solve: $$\sqrt{-x^2 + 10x - 16} < x - 2$$ 77. **Step 7:** For right side $$x - 2 \geq 0$$, i.e., $$x \geq 2$$, square both sides: $$-x^2 + 10x - 16 < (x - 2)^2$$ 78. **Step 8:** Expand right side: $$(x - 2)^2 = x^2 - 4x + 4$$ 79. **Step 9:** Inequality: $$-x^2 + 10x - 16 < x^2 - 4x + 4$$ 80. **Step 10:** Rearrange: $$-x^2 + 10x - 16 - x^2 + 4x - 4 < 0$$ 81. **Step 11:** Simplify: $$-2x^2 + 14x - 20 < 0$$ 82. **Step 12:** Multiply by -1 (flip inequality): $$2x^2 - 14x + 20 > 0$$ 83. **Step 13:** Divide by 2: $$x^2 - 7x + 10 > 0$$ 84. **Step 14:** Factor: $$(x - 5)(x - 2) > 0$$ 85. **Step 15:** Inequality holds for: $$x < 2$$ or $$x > 5$$ 86. **Step 16:** Combine with domain $$2 \leq x \leq 8$$ and $$x \geq 2$$: $$x > 5$$ and $$x \leq 8$$ 87. **Step 17:** Integral values: $$x = 6, 7, 8$$ 88. **Step 18:** Number of integral values is 3. --- 89. **Problem 13:** Find all possible values of $$f(x) = \frac{1 - x^2}{x^2 + 3}$$ 90. **Step 1:** Denominator $$x^2 + 3 > 0$$ for all real $$x$$. 91. **Step 2:** Rewrite function: $$f(x) = \frac{1 - x^2}{x^2 + 3} = \frac{1 + (-x^2)}{x^2 + 3}$$ 92. **Step 3:** Let $$t = x^2 \geq 0$$, then $$f(t) = \frac{1 - t}{t + 3}$$ 93. **Step 4:** Domain for $$t$$ is $$[0, \infty)$$. 94. **Step 5:** Find range of $$f(t)$$ for $$t \geq 0$$. 95. **Step 6:** Analyze limits: - As $$t \to 0$$, $$f(0) = \frac{1 - 0}{0 + 3} = \frac{1}{3}$$ - As $$t \to \infty$$, $$f(t) \to \frac{-t}{t} = -1$$ 96. **Step 7:** Check if $$f(t)$$ can reach $$-1$$: $$f(t) = -1 \Rightarrow \frac{1 - t}{t + 3} = -1$$ 97. **Step 8:** Multiply both sides: $$1 - t = -1(t + 3) = -t - 3$$ 98. **Step 9:** Rearrange: $$1 - t = -t - 3 \Rightarrow 1 = -3$$ contradiction. 99. **Step 10:** So $$f(t) \neq -1$$ but approaches it asymptotically. 100. **Step 11:** Find critical points by derivative or check monotonicity: 101. **Step 12:** Derivative: $$f'(t) = \frac{-(t + 3) - (1 - t)}{(t + 3)^2} = \frac{-t - 3 - 1 + t}{(t + 3)^2} = \frac{-4}{(t + 3)^2} < 0$$ 102. **Step 13:** Function strictly decreasing on $$[0, \infty)$$. 103. **Step 14:** Range is: $$(-1, \frac{1}{3}]$$ --- **Final answers:** 3. $$x \in (-6, -5] \cup [3, 4) \cup (4, 7]$$ 6. $$x \in (-2, \infty)$$ 7. $$x \in \mathbb{R} \setminus \{0, 2\}$$ 8. $$-\frac{9}{2} \leq x \leq 3$$ 9. $$x \in (-\infty, -2] \cup \{1\}$$ 10. $$x \in [-2, 2]$$ 11. $$x \in [-1, 2]$$ 12. Number of integral values = 3 13. $$y \in (-1, \frac{1}{3}]$$