Correct Answer: C\n\nTo solve the inequality $\\frac{\\sqrt{x^2 - 7x + 10}}{x^2 - 16} \\le 0$, we must consider several conditions for the expression to be valid and satisfy the inequality.\n\n**Step 1: Determine the domain for the square root expression.**\nFor the term $\\sqrt{x^2 - 7x + 10}$ to be a real number, the expression under the square root must be non-negative:\n$x^2 - 7x + 10 \\ge 0$\nFactorizing the quadratic expression:\n$(x-2)(x-5) \\ge 0$\nThis inequality holds when $x \\le 2$ or $x \\ge 5$. Thus, $x \\in (-\\infty, 2] \\cup [5, \\infty)$. (Let's call this **Condition A**)\n\n**Step 2: Determine restrictions on the denominator.**\nThe denominator of a fraction cannot be zero. Therefore:\n$x^2 - 16 \\ne 0$\n$(x-4)(x+4) \\ne 0$\nThis implies $x \\ne 4$ and $x \\ne -4$. (Let's call this **Condition B**)\n\n**Step 3: Analyze the overall inequality.**\nWe have the inequality $\\frac{\\text{Numerator}}{\\text{Denominator}} \\le 0$.\nSince the numerator, $\\sqrt{x^2 - 7x + 10}$, is a square root, it is inherently non-negative ($\\ge 0$) whenever it is defined. For the entire fraction to be less than or equal to zero, we must consider two distinct scenarios:\n\n**Scenario 1: The numerator is zero.**\nIf $\\sqrt{x^2 - 7x + 10} = 0$, then $x^2 - 7x + 10 = 0$, which implies $(x-2)(x-5) = 0$.\nSo, $x=2$ or $x=5$.\nWe must verify if these values satisfy **Condition B** (denominator not zero):\n* For $x=2$: Denominator is $2^2 - 16 = 4 - 16 = -12$. Since $-12 \\ne 0$, $x=2$ is a valid solution.\n* For $x=5$: Denominator is $5^2 - 16 = 25 - 16 = 9$. Since $9 \\ne 0$, $x=5$ is a valid solution.\nTherefore, $x=2$ and $x=5$ are two integer solutions.\n\n**Scenario 2: The numerator is positive AND the denominator is negative.**\n* **Numerator positive:** $\\sqrt{x^2 - 7x + 10} > 0$\n This implies $x^2 - 7x + 10 > 0$, so $(x-2)(x-5) > 0$.\n This inequality holds when $x 5$. Thus, $x \\in (-\\infty, 2) \\cup (5, \\infty)$. (Let's call this **Condition C**)\n\n* **Denominator negative:** $x^2 - 16 < 0$\n Factorizing: $(x-4)(x+4) < 0$.\n This inequality holds for $x$ values between the roots, i.e., $-4 < x < 4$. Thus, $x \\in (-4, 4)$. (Let's call this **Condition D**)\n\nTo satisfy Scenario 2, we need to find the intersection of **Condition C** and **Condition D**:\n$((-\\infty, 2) \\cup (5, \\infty)) \\cap (-4, 4)$\n* The intersection of $(-\\infty, 2)$ and $(-4, 4)$ is $(-4, 2)$.\n* The intersection of $(5, \\infty)$ and $(-4, 4)$ is an empty set ($\\emptyset$).\nTherefore, the solution from Scenario 2 is $x \\in (-4, 2)$.\n\nWe must also ensure that the values obtained from Scenario 2 satisfy **Condition A** (domain of the square root). The interval $(-4, 2)$ is a subset of $(-\\infty, 2]$, so these values are consistent with Condition A.\n\n**Step 4: Combine all valid solutions.**\nThe complete set of solutions for $x$ is the union of solutions from Scenario 1 and Scenario 2:\nSolution set $= \{2, 5\} \\cup (-4, 2)$.\nThis combined set can be more compactly written as $(-4, 2] \\cup \{5\}$.\n\n**Step 5: Count the number of integers in the solution set.**\nIntegers in the interval $(-4, 2]$ are: $-3, -2, -1, 0, 1, 2$. There are 6 integers in this range.\nThe integer $5$ is also a solution from Scenario 1.\n\nTotal number of integers satisfying the inequality is $6 + 1 = 7$.\n\nThe final answer is $\\boxed{7}$.