answer:Solution: For the function (f(x)=x^{3}+ax^{2}+(a+6)x+1), its derivative is (f′(x)=3x^{2}+2ax+(a+6)). Since the function has both a maximum and a minimum value, the equation (f′(x)=0) must have two distinct real roots. This means (3x^{2}+2ax+(a+6)=0) has two distinct real roots. Therefore, (Delta > 0), Thus, ((2a)^{2}-4×3×(a+6) > 0), solving this gives: (a < -3) or (a > 6). Hence, the answer is: boxed{a < -3 text{ or } a > 6}. First, we find the derivative of the function. Knowing that the function has both a maximum and a minimum value implies that the equation obtained by setting the derivative to (0) has two distinct real roots. By ensuring (Delta > 0), we can determine the range of (a). This problem uses the concept of extrema of a function to explore the application of derivatives in finding the extrema of functions. Transforming the condition that the function has both a maximum and a minimum value into the condition that the equation (f′(x)=0) has two distinct real roots is key to solving the problem.

answer:To analyze the given problems, we classify and compare the geometric properties of parallel and skew lines, as well as parallel and intersecting lines. Below are the similarities and differences for each pair of line relationships. Problem 1: Parallel Lines vs. Skew Lines Similarities: 1. **Coplanarity:** Neither pair is coplanar with the other. 2. **Non-intersecting:** Both parallel lines and skew lines do not intersect. Differences: 1. **Plane Definition:** - **Parallel Lines:** Parallel lines lie within the same plane. - **Skew Lines:** Skew lines lie in different planes (non-coplanar). 2. **Distance Preservation:** - **Parallel Lines:** The distance between any two points on parallel lines remains constant. - **Skew Lines:** There is no fixed distance between points on skew lines since they lie in different planes. Problem 2: Parallel Lines vs. Intersecting Lines Similarities: 1. **Straightness:** Both sets of lines are straight. Differences: 1. **Distance Preservation:** - **Parallel Lines:** Among parallel lines, the distance between them remains the same at all points. - **Intersecting Lines:** The distance between intersecting lines varies; at the point of intersection, the distance is zero. 2. **Angle Formation:** - **Parallel Lines:** Do not form angles with each other. - **Intersecting Lines:** Form angles at the point of intersection which can vary from 0 to 180 degrees. Conclusion: We have identified the various properties that distinguish and liken parallel lines to skew lines and intersecting lines. By understanding these properties, we can uniquely identify each relationship. boxed{text{Conclusion}}

answer:Given ( x star 48 = 3 ), transforming the expression: [ x star 48 = frac{sqrt{x + 48}}{sqrt{x - 48}} = 3 ] To solve for ( x ), square both sides: [ left(frac{sqrt{x + 48}}{sqrt{x - 48}}right)^2 = 3^2 ] [ frac{x + 48}{x - 48} = 9 ] Cross-multiply: [ x + 48 = 9(x - 48) ] [ x + 48 = 9x - 432 ] [ 480 = 8x ] [ x = 60 ] So, ( x = boxed{60} ).

answer:1. Consider a regular hexagon S with side length 1. Let S be the set of all points inside and on the boundary of the hexagon. 2. We need to find the smallest r > 0 such that any 3-coloring of S satisfies the condition: the distance between any two points of the same color is less than r. 3. Name the vertices of the hexagon in counterclockwise order as A, B, C, D, E, F. Let P, Q, R be the midpoints of segments AF, BC, DE respectively, and let O be the center of the hexagon. 4. Consider coloring the polygons ABQOP, OQCDR, and OREFP and their boundaries with three different colors: red, yellow, and blue as follows: - Segment OP is colored red. - Segment OQ is colored yellow except point O. - Segment OR is colored blue except point O. 5. It is evident that the distance between any two points of the same color is at most frac{3}{2}. Hence, the minimum r leq frac{3}{2}. 6. We will now prove that the smallest value of r is indeed r=frac{3}{2}. 7. Suppose, for contradiction, that the smallest value r_0 < frac{3}{2}. Let [ varepsilon = frac{1}{2} left( frac{3}{2} - r_0 right) > 0. ] 8. Choose a sufficiently large natural number n such that frac{1}{n} < varepsilon. 9. Divide each side of the given hexagon into n equal parts, and denote the 6n points on the boundary in order as M_0, M_1, M_2, ldots, M_{6n-1} where: [ begin{array}{l} A = M_0, B = M_n, C = M_{2n}, D = M_{3n}, E = M_{4n}, F = M_{5n}. end{array} ] and for j geq 6n, let M_j = M_{j-6n}. 10. Consider any k. Examine the set of three points {M_k, M_{2n+k}, M_{4n+k} }. It is easy to see that the distance between any two points among these three points is at least frac{3}{2} > r_0. Thus, these three points must all be of different colors. 11. Now, consider the set of three points {M_{k+1}, M_{2n+k}, M_{4n+k} }. We have: [ begin{array}{l} d(M_{k+1}, M_{2n+k}) geq d(M_k, M_{2n+k}) - d(M_{k+1}, M_k) geq frac{3}{2} - varepsilon = frac{3}{2} - frac{1}{2}left( frac{3}{2} - r_0 right) = frac{1}{2} left( frac{3}{2} + r_0 right) > r_0. end{array} ] Similarly, we have d(M_{k+1}, M_{4n+k}) > r_0. 12. Therefore, these three points must also all be of different colors. 13. Consequently, points M_k and M_{k+1} must be of the same color. 14. Due to the arbitrary nature of k, this implies that among the 6n points, there must be at least two points of the same color, leading to a contradiction. 15. Thus, the minimum value of r is: [ boxed{frac{3}{2}} ]