answer:(1) From the given equation bcos A + sqrt {3}bsin A - c - a = 0, we can use the Law of Sines, which implies: frac{b}{sin B} = frac{a}{sin A} = frac{c}{sin C}. Multiplying through by sin B, we get: bcos A + sqrt {3}bsin A - sin Csin B - sin Asin B = 0. Considering that sin C = sin (A + B) due to the angle sum in a triangle and using the sine of sum identity, we get: sin Bcos A + sqrt {3}sin Bsin A - sin Acos B - cos Asin B - sin A = 0. After simplifying, we have: sqrt {3}sin Bsin A - sin Acos B - sin A = 0. Given sin A neq 0, we can divide the whole equation by sin A, yielding: sqrt {3}sin B - cos B = 1. By rewriting the equation as: sin(B - frac {pi}{6}) = frac {1}{2}, And since B is in the interval (0, pi), we have: B = frac {pi}{3}. (2) Using the cosine rule from (1), we have: b^2 = a^2 + c^2 - 2accos B. Substituting b = sqrt {3} into the equation, we get: a^2 + c^2 - ac = 3. Thus, the square of a+c is: (a+c)^{2} = 3 + 3ac. By the AM-GM inequality, the product ac can be maximized when a = c. Hence: (a+c)^{2} = 3 + 3ac leq 3 + 3( frac {a+c}{2})^{2}, Setting the right side to be an equality, we solve to get that a+c is maximized when: a+c leq 2 sqrt {3}. Therefore, the maximum value of a+c is achieved when a = c = sqrt {3}, and the maximum value is: a+c = boxed{2 sqrt {3}}.

answer:Ni.7.p0. Thus, p>1 or p<-1.Let the two real roots of the equation x^{2}+2 p x+1=0 be x_{1} and x_{2}, then we havex_{1}+x_{2}=-2 p, x_{1} x_{2}=1 .Since left(x_{1}-1right)left(x_{2}-1right)<0, it follows that,x_{1} x_{2}-left(x_{1}+x_{2}right)+1<0 text {. }Thus, 2 p+2<0, which means p<-1.

answer:Solution: (1) (-14) - 14 + (-5) - (-30) - (+2) = -14 - 14 - 5 + 30 - 2 = -(14 + 14 + 5 + 2) + 30 = -35 + 30 = -5; So, the answer is boxed{-5}.(2) -1^{2008} - [5 times (-2) - (-4)^2 div (-8)] = -1 - [-10 - 16 div (-8)] = -1 - [-10 + 2] = -1 + 8 = 7. Thus, the answer is boxed{7}.

answer:BTranslate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.

answer:(1) left[0, frac{sqrt{6}}{6}right]

answer:Answer: 2^{n-1}.Let's denote the numbers of our permutation from left to right as a_{1}, a_{2}, ldots, a_{n}.Solution 1. We will start from the end. The last number a_{n} of a funny permutation is either greater or less than all the numbers in the set 1,2,3, ldots, n, so it is either 1 or n. The second-to-last number a_{n-1} of a funny permutation is either greater or less than all the numbers in the set 1,2,3, ldots, n except a_{n}, i.e., it is the smallest or largest element in the set 1,2,3, ldots, n-1 or in the set 2,3, ldots, n. In each case, there are exactly two choices, and the options for the last two numbers of the permutation look like this: (n-1, n), (1, n), (2,1), (n, 1). It is not difficult to verify that for any k=n, n-1, ldots, 2,1, the first k numbers a_{1}, a_{2}, ldots, a_{k} of the permutation form an interval of k consecutive numbers from the set 1,2,3, ldots, n, and the number a_{k} is the minimum or maximum in this interval—there are two possibilities, except for the very first number a_{1}, for which only one possibility remains. In total, we get exactly 2^{n-1} possible choices.Grading criteria for Solution 1. (cdot) Noted and justified that the last number of a funny permutation is 1 or n: 1 point.(cdot) Noted and explicitly stated that the set a_{1}, a_{2}, ldots, a_{k} forms an interval of k consecutive numbers from the set 1,2,3, ldots, n: 1 point.(cdot) Stated that there are exactly two choices for each subsequent a_{k}: 1 point.(cdot) Specified that a_{k} is chosen as the maximum or minimum of the remaining interval: 1 point.(cdot) From this, concluded that there are exactly 2^{n-1} possible choices for the permutation: 3 points.Solution 2. Let a_{1}=m, where m is one of the numbers 1,2, cdots, n. Any number a_{k} less than m will, by condition, also be less than all numbers less than m and standing to the left of a_{k}, so all numbers in the funny permutation less than m form a decreasing subsequence. Similarly, all numbers greater than m form an increasing subsequence. Any mutual arrangement of these subsequences satisfies the condition and leads to a funny permutation.Therefore, any funny permutation is completely determined by the value of its first element m=1,2, ldots, n and the positions of m-1 places where the numbers m-1, m-2, ldots, 1 are placed in decreasing order from left to right among the n-1 members of the funny permutation, except the first. The remaining places are automatically filled with the numbers m+1, m+2, ldots, n in ascending order. Therefore, for a fixed m=1,2, ldots, n, the number of funny permutations is the number of combinations C_{n-1}^{m-1}, and the total number of them is the sum C_{n-1}^{0}+C_{n-1}^{1}+ldots+C_{n-1}^{n-1}=2^{n-1}.Grading criteria for Solution 2. (cdot) Noted and justified that all numbers less than m form a decreasing subsequence: 1 point.(cdot) Noted and justified that all numbers greater than m form an increasing subsequence: 1 point. (cdot) If either or both of the previous points lack justification: minus 1 point.(cdot) Stated (1 point) and proved (1 point) that any mutual arrangement of these subsequences leads to a funny permutation: total 2 points.(cdot) Stated that any funny permutation is completely determined by the choice of its first element and the positions of the numbers less than the first, placed in decreasing order from left to right among all members of the funny permutation: 1 point.(cdot) Correctly indicated that for a fixed m=1,2, ldots, n, the number of funny permutations is C_{n-1}^{m-1}: 1 point.(cdot) Correctly found the sum C_{n-1}^{0}+C_{n-1}^{1}+ldots+C_{n-1}^{n-1}=2^{n-1}: 1 point.Solution 3. We will start from the beginning, considering how the set A_{k}=left{a_{1}, a_{2}, ldots, a_{k}right} of the first k numbers of a funny permutation changes as k=1,2,3, ldots, n increases. As a_{1}, we can take any natural number from the interval 1,2,3, ldots, n, and A_{1}=left{a_{1}right}. Suppose that on the k-th step, the numbers a_{1}, a_{2}, ldots, a_{k} have already been chosen, and let m_{k} and M_{k} be the minimum and maximum of them, respectively.We will prove that the next a_{k+1} must be either m_{k}-1 or M_{k}+1. Indeed, if m_{k} < M_{k}, then a_{k+1} must be either m_{k}-1 or M_{k}+1 because any other choice would violate the condition of the funny permutation. If m_{k} > M_{k}, this is impossible, so on each step, the difference between M_{k} and m_{k} increases by at least 1. The number of steps is n-1, and the final difference between M_{n} and m_{n} does not exceed n-1, so on each step, it increases by exactly 1. From this, it immediately follows that for each k=1,2,3, ldots, n, the set A_{k} consists of some k consecutive numbers from the interval 1,2,3, ldots, n, and A_{k+1} is obtained by adding a number adjacent to this interval on the left or right. The