answer:To solve for ( f(3) ), substitute ( x = 3 ) into the function: [ f(3) = dfrac{3^2 + 3 + 1}{3(3^2) - 4} ] [ = dfrac{9 + 3 + 1}{3(9) - 4} ] [ = dfrac{13}{27 - 4} ] [ = dfrac{13}{23} ] Thus, the value of ( f(3) ) is: [ boxed{dfrac{13}{23}} ]

answer:1. **Function Relationship Establishment**: Setting up the relation gleft(frac{x}{2}right) = x^2 - 4x + 3, we need to find all k such that g(2k) = 6. 2. **Relating g(2k) to the given function**: Using x = 4k (since frac{4k}{2} = 2k), we substitute: [ g(2k) = gleft(frac{4k}{2}right) = (4k)^2 - 4 times 4k + 3 = 16k^2 - 16k + 3 ] 3. **Setting the equation**: Setting g(2k) = 6 gives us: [ 16k^2 - 16k + 3 = 6 ] Simplifying, [ 16k^2 - 16k - 3 = 0 ] 4. **Using the Quadratic Formula**: The roots of the quadratic equation can be found as follows: [ k = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] Where a = 16, b = -16, and c = -3, [ k = frac{16 pm sqrt{(-16)^2 - 4 times 16 times (-3)}}{2 times 16} ] [ k = frac{16 pm sqrt{256 + 192}}{32} ] [ k = frac{16 pm sqrt{448}}{32} ] [ k = frac{16 pm 8sqrt{7}}{32} ] Simplifying, [ k_1 = frac{1}{2} + frac{sqrt{7}}{4}, quad k_2 = frac{1}{2} - frac{sqrt{7}}{4} ] 5. **Sum of the Roots**: The sum is simply: [ left(frac{1}{2} + frac{sqrt{7}}{4}right) + left(frac{1}{2} - frac{sqrt{7}}{4}right) = 1 ] 6. **Conclusion**: The sum of all values of k for which g(2k) = 6 is 1. The correct answer is boxed{(A) , 1}.

answer:To solve this problem, we can use the concept of man-days, which is a way to quantify the amount of work done by a certain number of workers over a certain number of days. First, let's calculate the total man-days required to complete the work based on the information given for 12 men: Total man-days for 12 men = 12 men * 80 days = 960 man-days Now, let's calculate the total man-days required to complete the work based on the information given for 16 men: Total man-days for 16 men = 16 men * 60 days = 960 man-days We can see that the total man-days required to complete the work is the same in both cases (960 man-days), which makes sense because the amount of work doesn't change regardless of the number of men. Now, let's find out how many days it would take for a different number of men, say "x" men, to complete the work. We can set up the following equation: x men * y days = 960 man-days We want to find "y" (the number of days it would take for "x" men to complete the work). To do this, we can rearrange the equation to solve for "y": y days = 960 man-days / x men Since we don't have a specific number for "x" men, we can't calculate the exact number of days "y." However, this equation shows us how to find the number of days it would take for any given number of men to complete the work, as long as we know the total man-days required boxed{(960} in this case).

answer:(I) When a=2, the function becomes f(x)=2ln x - x^2 + 2x. To find the equation of the tangent we first need the derivative of f(x): [ f'(x) = frac{2}{x} - 2x + 2. ] The coordinates of the point of tangency are (1, f(1))=(1, 1). The slope of the tangent (denoted by k) at x=1 is: [ k=f'(1)=frac{2}{1} - 2 cdot 1 + 2 = 2. ] Thus, the equation of the tangent line to the graph of f(x) at x=1 is: [ y - 1 = 2(x - 1), quad text{which simplifies to} quad y = 2x - 1. ] (II) The function g(x) is given by: [ g(x) = f(x) - ax + m = 2ln x - x^2 + m. ] We find its derivative: [ g'(x) = frac{2}{x} - 2x = frac{-2(x^2-1)}{x}. ] Since x in left[frac{1}{e}, eright], setting g'(x) = 0 gives us x = 1. For frac{1}{e} < x < 1, we have g'(x) > 0, which means g(x) is increasing. For 1 < x < e, we have g'(x) < 0, which means g(x) is decreasing. Therefore, g(x) has a local maximum at x = 1, which is: [ g(1) = m - 1. ] The value of g(x) at the endpoints of the interval is: [ gleft(frac{1}{e}right) = m - 2 - frac{1}{e^2} quad text{and} quad g(e) = m + 2 - e^2. ] Since g(e) - gleft(frac{1}{e}right) = 4 - e^2 + frac{1}{e^2} < 0, we have g(e) < gleft(frac{1}{e}right). For g(x) = f(x) - ax + m to have two zeros on left[frac{1}{e}, eright], we need: [ begin{cases} g(1) = m - 1 > 0, gleft(frac{1}{e}right) = m - 2 - frac{1}{e^2} leq 0. end{cases} ] Solving these inequalities, we get: [ 1 < m leq 2 + frac{1}{e^2}. ] Hence, the range of the real number m is boxed{left(1, 2 + frac{1}{e^2}right]}.