answer:Answer: A: -( -2010 ) = 2010, which is a positive number, so this option is incorrect; B: -| -2010 | = -2010, which is a negative rational number, so this option is correct; C: (-2011)^{2010} = 2011^{2010}, which is a positive number, so this option is incorrect; D: frac{-2010}{-2011} = 1, which is a positive number, so this option is incorrect. Therefore, the correct answer is boxed{text{B}}.

answer:Given a_1=1, a_{n+1}= frac {a_n}{a_n+2}(n∈N^{*}), we can find that:a_2= frac {1}{1+2}= frac {1}{3},a_3= frac { frac {1}{3}}{ frac {1}{3}+2}= frac {1}{7},a_4= frac { frac {1}{7}}{ frac {1}{7}+2}= frac {1}{15}.Thus, we can conjecture that a_n= frac {1}{2^{n}-1}.We will prove this by mathematical induction:1. When n=1, a_1= frac {1}{2^{1}-1}=1, which holds true.2. Assume that when n=k, the equation holds true, i.e., a_k= frac {1}{2^{k}-1}.Then, when n=k+1, we have a_{k+1}= frac {a_k}{a_k+2}= frac { frac {1}{2^{k}-1}}{ frac {1}{2^{k}-1}+2}= frac {1}{2^{k+1}-1}, which also holds true.Therefore, a_n= frac {1}{2^{n}-1}.Consequently, b_{n+1}=(n-λ)( frac {1}{a_n}+1)=(n-λ)⋅2^{n}, and b_2=(1-λ)⋅2=2-2λ.Given that b_1=-λ, and the sequence {b_n} is monotonically increasing, we have b_1=-λ < b_2=2-2λ.Solving for λ, we obtain λ < 2.Hence, the answer is boxed{C}.This problem tests the method of finding the general term formula of a sequence and its application. When solving the problem, carefully read the question, answer it meticulously, and pay attention to the reasonable use of mathematical induction and equivalent transformation thinking.

answer:There are two cases:- If p mid n, then we can write n=k p with k in mathbb{N}^{*}. We then have that k p(k p+p)=k(k+1) p^{2} is a perfect square, so k(k+1) is a perfect square. Since k and k+1 are coprime, each one is a perfect square. Writing k=a^{2} and k+1=b^{2} with a, b>0, we get 1=a^{2}-b^{2}=(a+b)(a-b). Since k geq 1, a geq 1 and b geq 1, 1 has an integer divisor, which is impossible.- Otherwise, n and n+p are coprime, so both are squares. We write n=a^{2} and n+p=b^{2} with a, b>0. Then (b-a)(b+a)=b^{2}-a^{2}=p. Since p is prime and b+a geq 2 divides p, we have a+b=p and b-a=1, so b=frac{p+1}{2} and a=frac{p-1}{2}, hence n=left(frac{p-1}{2}right)^{2} and p is odd.Conversely, the pairs left(left(frac{p-1}{2}right)^{2}, pright) with p an odd prime work because:left(frac{p-1}{2}right)^{2}left(left(frac{p-1}{2}right)^{2}+pright)=left(frac{p^{2}-1}{2}right)^{2}

answer:31. Ans: 8begin{aligned}& x sqrt{y}+y sqrt{x}+sqrt{2006 x y}-sqrt{2006 x}-sqrt{2006 y}-2006=0 Longleftrightarrow & (sqrt{x}+sqrt{y}+sqrt{2006})(sqrt{x y}-sqrt{2006})=0 Longleftrightarrow & sqrt{x y}-sqrt{2006}=0 quad(text { since } sqrt{x}+sqrt{y}+sqrt{2006}>0) Longleftrightarrow & x y=2006=17 times 59 times 2 .end{aligned}Thus the solutions are(1,2006),(2006,1),(2,1003),(1003,2),(34,59),(59,34),(17,118),(118,17) .

answer:Answer: (2,2,2) and (3,4,3).Solution 1. Clearly, a>1. We consider three cases.Case 1: We have ab which is also impossible since in this case we have b!leqslant a!a>1.Case 2: We have a>p. In this case b!-a^{p}-p>p^{p}-p geqslant p! so b>p which means that a^{p}-b!+p is divisible by p. Hence, a is divisible by p and b!-a^{p}-p is not divisible by p^{2}. This means that b(2 p-1)!+p geqslant b!+p.Comment. The inequality p^{2 p}>(2 p-1)!+p can be shown e.g. by using(2 p-1)!=[1 cdot(2 p-1)] cdot[2 cdot(2 p-2)] cdots cdots[(p-1)(p+1)] cdot pp! and so b geqslant p+1 which implies thatv_{2}((p+1)!) leqslant v_{2}(b!)-v_{2}left(p^{p-1}-1right) stackrel{L T E}{ } 2 v_{2}(p-1)+v_{2}(p+1)-1-v_{2}left(frac{p-1}{2} cdot(p-1) cdot(p+1)right),where in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of (p+1)!. But, due to p+1 geqslant 8, there are at least 4 even numbers among 1,2, ldots, p+1, so this case is not possible.Solution 2. The cases a neq p are covered as in solution 1 , as are p-2,3. For p geqslant 5 we have b!-pleft(p^{p-1}-1right). By Zsigmondy's Theorem there exists some prime q that divides p^{p-1}-1 but does not divide p^{k}-1 for k(2 p-1)^{p-1} p>p^{p}>p^{p}-p,a contradiction.Solution 3. The cases a neq p are covered as in solution 1, as are p-2,3. Also b>p, as p^{p}>p!+p for p>2. The cases p-5,7,11 are also checked manually, so assume p geqslant 13. Let q mid p+1 be an odd prime. By LTEv_{q}left(p^{p}-pright)-v_{q}left(left(p^{2}right)^{frac{p-1}{2}}-1right)-v_{q}left(p^{2}-1right)+v_{q}left(frac{p-1}{2}right)-v_{q}(p+1) .But b geqslant p+1, so then v_{q}(b!)>v_{q}(p+1), since qq^{d-1} geqslant 2 dprovided d geqslant 2 and q>3, or d geqslant 3.If q-3, d-2 and p geqslant 13 then v_{q}(b!) geqslant v_{q}(p!) geqslant v_{q}(13!)-5>2 d. Either way, d leqslant 1.If p>2 q+1 (so p>3 q, as q mid p-1 ) thenv_{q}(b!) geqslant v_{q}((3 q)!)-3,so we must have q geqslant frac{p}{2}, in other words, p-1-2 q. This implies that p-2^{k}-1 and q-2^{k-1}-1 are both prime, but it is not possible to have two consecutive Mersenne primes.Solution 4. Let a-p, b>p and p geqslant 5 (the remaining cases are dealt with as in solution 3). Modulo (p+1)^{2} it holds thatp^{p}-p-(p+1-1)^{p}-p=binom{p}{1}(p+1)(-1)^{p-1}+(-1)^{p}-p-p(p+1)-1-p-p^{2}-1 neq 0 quad bmod left((p+1)^{2}right) text {. }Since p geqslant 5, the numbers 2 and frac{p+1}{2} are distinct and less than or equal to p. Therefore, p+1 mid p!, and so (p+1)^{2} mid(p+1)!.But b geqslant p+1, so b!=0 neq p^{p}-p bmod (p+1)^{2}, a contradiction.

answer:Since C_n^6 = C_n^4, we have n=10. Therefore, (2x-5)^{10} = a + a_1(x-1) + a_2(x-1)^2 + ldots + a_n(x-1)^{10}, By setting x=2, we get 1 = a + a_1 + a_2 + ldots + a_n. Hence, the correct choice is boxed{A}.