answer:Solution.Domain of definition: -1<p<1.begin{aligned}& left(left(1-p^{2}right)^{-1 / 2}-left(1+p^{2}right)^{-1 / 2}right)^{2}+2left(1-p^{4}right)^{-1 / 2}=left(frac{1}{sqrt{1-p^{2}}}-frac{1}{sqrt{1+p^{2}}}right)^{2}+ & +frac{2}{sqrt{1-p^{4}}}=left(frac{sqrt{1+p^{2}}-sqrt{1-p^{2}}}{sqrt{1-p^{4}}}right)^{2}+frac{2}{sqrt{1-p^{4}}}=frac{1+p^{2}-2 sqrt{1-p^{4}}+1-p^{2}}{1-p^{4}}+ & +frac{2}{sqrt{1-p^{4}}}=frac{2-2 sqrt{1-p^{4}}}{1-p^{4}}+frac{2}{sqrt{1-p^{4}}}=frac{2-2 sqrt{1-p^{4}}+2 sqrt{1-p^{4}}}{1-p^{4}}=frac{2}{1-p^{4}}end{aligned}Answer: frac{2}{1-p^{4}}.

answer:(1) From 2S= overrightarrow {AB}cdot overrightarrow {AC}, we obtain bc sin A = bc cos A. Since A in (0, pi), we have tan A = 1, which leads to A = boxed{frac{pi}{4}}.(Note: The angle A is in the interval (0, pi) because it's an angle within a triangle, and the sum of the angles of a triangle is pi.)(2) In triangle ABC, we have cos B = frac{4}{5}, so sin B = frac{3}{5}. Now, we'll find sin C:sin C = sin(A + B) = sin A cos B + cos A sin B = frac{7sqrt{2}}{10}.Applying the sine law, frac{a}{sin A} = frac{c}{sin C}, we have:frac{a}{frac{sqrt{2}}{2}} = frac{7}{frac{7sqrt{2}}{10}}Solving for a, we find a = boxed{5}.(Note: The sine law, also known as the law of sines, states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.)

answer:begin{aligned}-Delta_{1} & =(-8 n)^{2}-4 times 4 times(-3 n-2) & =(8 n+3)^{2}+23>0 .end{aligned}Then n is any real number, the first equation always has real roots.Let the two roots of the first equation be alpha, beta, thenbegin{array}{l}alpha+beta=2 n, alpha beta=frac{-3 n-2}{4} . therefore(alpha-beta)^{2}=(alpha+beta)^{2}-4 alpha beta=4 n^{2}+3 n+2 .end{array}From the second equation, we get[x-(2 n+2)][x+(n-1)]=0 text {. }Solving, we get x_{1}=2 n+2, x_{2}=1-n.If x_{1} is an integer, then4 n^{2}+3 n+2=2 n+2 text {. }Solving, we get n_{1}=0, n_{2}=-frac{1}{4}.When n=0, x_{1}=2;When n=-frac{1}{4}, x_{1}=frac{3}{2} is not an integer. Discard.If x_{2} is an integer, then 4 n^{2}+3 n+2=1-n.Solving, we get n_{3}=n_{4}=-frac{1}{2}.When n=-frac{1}{2}, x_{2}=frac{3}{2} is not an integer. Discard.In summary, when n=0, the square of the difference of the two real roots of the first equation equals the integer root of the second equation.

answer:3. We will consider the sum S=sum_{i=1}^{n}left|a_{i}-a_{i+1}right|, where a_{n+1}=a_{1}. This sum can be written in the formS=sum_{i=1}^{n} epsilon_{i}left(a_{i}-a_{i+1}right)=sum_{i=1}^{n} c_{i} a_{i} quad text { for some } quad epsilon_{i} in{-1,1} quad text { and } quad c_{i}=epsilon_{i}-epsilon_{i-1},where epsilon_{0}=epsilon_{n}. For each i, c_{i} in{-2,0,2}. Consider the sets A=left{i mid c_{i}=2right} and B=left{i mid c_{i}=-2right}. Since c_{1}+c_{2}+cdots+c_{n}=0, we have |A|=|B|=k for some k leqslant frac{n}{2}. Now we haveS=2 sum_{i in A} x_{i}-2 sum_{i in B} x_{i} leqslant 2[n+(n-1)+cdots+(n-k+1)]-2[1+2+cdots+k]=2 k(n-k)The expression 2 k(n-k) is maximized for k=left[frac{n}{2}right] and is then equal to left[frac{n^{2}}{2}right]. Finally, since left|a_{n}-a_{1}right| geqslant 1, we haveS leqslantleft[frac{n^{2}}{2}right]-1Equality is achieved, for example, for the permutation left(a_{1}, a_{2}, ldots, a_{n}right)=(m+1,1, m+2,2, m+3,3, ldots, m), where m=n-left[frac{n}{2}right].Second solution. The sum S=sum_{i=1}^{n}left|a_{i}-a_{i+1}right| represents the length ell of the closed broken line A_{1} A_{2} ldots A_{n} A_{1}, where A_{i} is a point on the real line with coordinate a_{i}.For each k=1, ldots, n-1, consider the unit segment [k, k+1]. Each of the n segments A_{i} A_{i+1} that covers it has one endpoint in the interval [1, k], and the other in the interval [k+1, ldots, n]. It follows that this segment is covered by at most 2 min {k, n-k} segments A_{i} A_{i+1}(1 leqslant i leqslant n). In total, the observed broken line covers at most 2 sum_{k=1}^{n-1} min {k, n-k}=left[frac{n^{2}}{2}right] unit segments, hence ell leqslantleft[frac{n^{2}}{2}right].Finally, due to left|a_{n}-a_{1}right| geqslant 1, the desired sum is not greater than left[frac{n^{2}}{2}right]-1.

answer:Since the set {x in mathbb{N}^{+} | x-3 < 2} is represented using the descriptive method, another way to represent it is by using the enumeration method. Since {x in mathbb{N}^{+} | x-3 < 2} = {x in mathbb{N}^{+} | x < 5} = {1, 2, 3, 4}, Therefore, the correct choice is: boxed{text{B}}. The set {x in mathbb{N}^{+} | x-3 < 2} is represented using the descriptive method, and another way to represent it is by using the enumeration method, which involves listing the elements of the set as described. This question examines the methods of representing sets, which is a basic topic. The key to solving the problem is to clearly understand the representation of the elements given in the problem, which are positive natural numbers.

answer:(I) Since -8a_1, a_2, a_3 form an arithmetic sequence, we have 2a_2=-8a_1+a_3. Substituting a_1=2 and a_n=a_1 q^{n-1}, we get 2 cdot 2q=-8 cdot 2+2q^2, which simplifies to q^2-2q-8=0. Since q > 0, the solution is q=4. Thus, the general term formula for the sequence {a_n} is a_n = 2 cdot 4^{n-1} = 2^{2n-1}.(II) Using b_n=log _2 a_n and a_n=2^{2n-1}, we get b_n=2n-1. Thus, frac{1}{b_n b_{n+1}} = frac{1}{(2n-1)(2n+1)} = frac{1}{2}(frac{1}{2n-1}-frac{1}{2n+1}). Therefore, the sum of the first n terms for the sequence {frac{1}{b_n b_{n+1}}} is:begin{align*}T_n &= frac{1}{2}left[left(1-frac{1}{3}right)+left(frac{1}{3}-frac{1}{5}right)+ldots+left(frac{1}{2n-1}-frac{1}{2n+1}right)right] &= frac{1}{2}left(1-frac{1}{2n+1}right) &= boxed{frac{n}{2n+1}}.end{align*}