answer:1. **Given**: ( p equiv 3 ,(textrm{mod}, 4) ) is a prime and (theta) is an angle such that (tan(theta)) is rational. We need to prove that (tan((p+1)theta)) is a rational number with numerator divisible by ( p ). 2. **Identity for (tan(nx))**: [ tan(nx) = frac{sumlimits_{k text{ odd}} (-1)^{frac{k-1}{2}} binom{n}{k} tan^k(x)}{sumlimits_{k text{ even}} (-1)^{frac{k}{2}} binom{n}{k} tan^k(x)} ] This implies that (tan((p+1)theta) in mathbb{Q}). Therefore, there exist polynomials (P_n) and (Q_n) such that: [ tan(nx) = frac{P_n(tan(x))}{Q_n(tan(x))} ] If (n) is even, (P_n) and (Q_n) have degrees (n-1) and (n) respectively. If (n) is odd, (P_n) and (Q_n) have degrees (n) and (n-1) respectively. 3. **Degrees of Polynomials**: For (n = p+1), since (p+1) is even, we have: [ operatorname{deg}(P_{p+1}) = p quad text{and} quad operatorname{deg}(Q_{p+1}) = p+1 ] 4. **Expressing (tan(theta))**: Write (tan(theta) = frac{u}{v}), where (u) and (v) are coprime integers. Assume (u, v > 0). 5. **Case 1: (p mid v)**: [ tan((p+1)theta) = frac{P_{p+1}left(frac{u}{v}right)}{Q_{p+1}left(frac{u}{v}right)} = frac{v cdot left(v^p P_{p+1}left(frac{u}{v}right)right)}{v^{p+1} Q_{p+1}left(frac{u}{v}right)} ] Here, the numerator is divisible by (v) (and thus by (p)), since (v^p P_{p+1}left(frac{u}{v}right)) and (v^{p+1} Q_{p+1}left(frac{u}{v}right)) are both polynomials in (mathbb{Z}[u,v]). The leading term of (Q_{p+1}) is (pm u^{p+1} notequiv 0 pmod{p}), so the denominator is not divisible by (p). 6. **Case 2: (p nmid v)**: [ frac{u}{v} text{ can be interpreted as a residue in } mathbb{F}_p ] We need to show (p mid frac{P_{p+1}(r)}{Q_{p+1}(r)}) for each (r in mathbb{F}_p). 7. **Möbius Transformation**: Define (x_i = tan(itheta)). Then: [ x_{i+1} = tan(itheta + theta) = frac{tan(itheta) + tan(theta)}{1 - tan(itheta) tan(theta)} = frac{x_i + r}{1 - r x_i} ] This is a Möbius transformation. Iterating this is equivalent to exponentiating the matrix: [ M = begin{pmatrix} 1 & r r & 1 end{pmatrix} ] 8. **Matrix Determinant**: Since (p equiv 3 pmod{4}), (det(M) = r^2 + 1 notequiv 0 pmod{p}). Thus, (H := {M^i mid i in mathbb{Z}}) is a subgroup of (G := operatorname{GL}_2(mathbb{F}_p)). 9. **Order of (M)**: The desired statement (x_{p+1} = 0) is equivalent to (operatorname{ord}(M) mid p+1). 10. **Subgroup (H')**: [ H' = left{ begin{pmatrix} a & b b & a end{pmatrix} mid a, b in mathbb{F}_p, a neq 0 lor b neq 0 right} ] This is a subgroup of (G) with order ((p-1)(p+1)). Since (H leq H'), by Lagrange's theorem, (operatorname{ord}(M) mid (p-1)(p+1)). 11. **Equivalence Relations**: Construct two equivalence relations: [ S = left{ begin{pmatrix} a b end{pmatrix} mid a, b in mathbb{F}_p, a neq 0 lor b neq 0 right} ] Define (ast) by: [ begin{pmatrix} a b end{pmatrix} ast begin{pmatrix} c d end{pmatrix} iff ad - bc = 0 ] The set (S/ast) of equivalence classes can be thought of as (mathbb{F}_p cup {infty}). 12. **Another Equivalence Relation**: Define: [ V bullet W iff exists v in V, w in W, i in mathbb{Z}: v = M^i w ] This is an equivalence relation since (H) is a group. The residue classes of (bullet) are of equal size (namely, (operatorname{ord}(M))), so (operatorname{ord}(M)) divides (|S/ast| = p+1). (blacksquare)

answer:For ((1)), the original inequality can be transformed into: ((x-6)(x+1) < 0), For ((2)), we have ((x-1)(x+2)leqslant 0) and (x+2neq 0), Therefore, the solution set for the inequality is (boxed{{x|-1 < x < 6}}) for the first inequality, and the solution set for the second inequality is (boxed{{x|-2 < x leqslant 1}}).

answer:: 1. Let (S = sum_{i=1}^{n} b_i a_i). 2. We can rewrite (S) by expanding the terms as follows: [ S = b_1 a_1 + sum_{i=2}^{n} b_i left( a_1 + ldots + a_i right) - sum_{i=2}^{n} b_i left( a_1 + ldots + a_{i-1} right) ] 3. Combining the sums, we get: [ S = sum_{i=1}^{n} left( a_1 + ldots + a_i right) b_i - sum_{i=1}^{n-1} left( a_1 + ldots + a_i right) b_{i+1} ] 4. Notice that: [ b_{n+1} = 0 ] 5. Thus, we can rewrite the previous equation as: [ S = sum_{i=1}^{n} left( a_1 + ldots + a_i right) (b_i - b_{i+1}) ] 6. Define (A_i = left| a_1 + ldots + a_i right| ) for (i = 1, ldots, n). 7. Among the numbers (A_1, A_2, ldots, A_n), select the largest one, and let this maximum be denoted by (A_k). 8. We then have: [ |S| leq sum_{i=1}^{n} left| a_1 + ldots + a_i right| cdot left| b_i - b_i+1 right| ] 9. Using the upper bound (A_k): [ |S| leq sum_{i=1}^{n} A_k left(b_i - b_{i+1}right) ] 10. Since (b_i geq b_{i+1}): [ sum_{i=1}^{n} left( b_i - b_{i+1} right) = b_1 - b_{n+1} = b_1 leq 1 ] 11. Therefore: [ |S| leq A_k sum_{i=1}^{n} left( b_i - b_{i+1} right) = A_k ] 12. Hence: [ |S| leq left| a_1 + ldots + a_k right| ] 13. By carefully choosing (k), it confirms that there exists such a (k) satisfying the inequality: [ left| sum_{i=1}^{n} b_i a_i right| leq left| sum_{i=1}^{n} a_i right| ] # Conclusion: [ boxed{} ]

answer:To find out how many seconds it takes for the train to cross the man, we need to convert the speed of the train from kilometers per hour (km/h) to meters per second (m/s), and then use the length of the train to calculate the time. First, let's convert the speed from km/h to m/s: [ text{Speed in m/s} = text{Speed in km/h} times frac{1000 text{ meters}}{1 text{ kilometer}} times frac{1 text{ hour}}{3600 text{ seconds}} ] [ text{Speed in m/s} = 80 times frac{1000}{1} times frac{1}{3600} ] [ text{Speed in m/s} = 80 times frac{1000}{3600} ] [ text{Speed in m/s} = frac{80000}{3600} ] [ text{Speed in m/s} = frac{20000}{900} ] [ text{Speed in m/s} = frac{2000}{90} ] [ text{Speed in m/s} = frac{200}{9} ] [ text{Speed in m/s} = 22.overline{22} ] Now that we have the speed in meters per second, we can calculate the time it takes for the train to cross the man using the length of the train: [ text{Time (in seconds)} = frac{text{Length of the train (in meters)}}{text{Speed of the train (in m/s)}} ] [ text{Time (in seconds)} = frac{200}{22.overline{22}} ] [ text{Time (in seconds)} = frac{200}{frac{200}{9}} ] [ text{Time (in seconds)} = 200 times frac{9}{200} ] [ text{Time (in seconds)} = 9 ] So, it takes boxed{9} seconds for the train to cross the man.