Đề thi ôlympic Sinh họcLớp 8

... them (may be the same) must be divisible by 5 2006 . Therefore, x must satisfy the following two conditions: x ≡ 0 or 1 (mod 2 2006 ); x ≡ 0 or 1 (mod 5 2006 ). Altogether we have 4 cases. The Chinese ... unique solution among the numbers 0, 1, . . . , 10 2006 − 1. These four numbers are different because each two gives different residues modulo 2 2006 or 5 2006 . Moreover, one of t...

Ngày tải lên: 17/10/2013, 23:15

... IMC2007, Blagoevgrad, Bulgaria Day 2, August 6, 2007 Problem 1. Let f : R → R be a continuous function. Suppose that for ... using only a translation or a rotation. Does this imply that f(x) = ax + b for some real numbers a and b ? Solution. No. The function f(x) = e x also has this property since ce x = e x+log c . Problem ... with a ij = 2 if i = j 1 if i − j ≡ ±2 (mod n) 0 otherwise...

Ngày tải lên: 20/10/2013, 18:15

... polynomial f (x) 2008. Then f(a i ) = g(a i )h(a i ) = 2008 for i = 1, . . . , 81, Hence, g(a 1 ), . . . , g(a 81 ) are integer divisors of 2008. Since 2008 = 2 3 ·251 (2, 251 are primes) then 2008 has ... orthonormal system. This it true for any distinct x ′ , x ′′ ∈ S \ T ; it follows that the entire system  √ 2 d (x − y) : x ∈ S  is an orthonormal system of vectors in H, as requi...

Ngày tải lên: 20/10/2013, 18:15

... VnMath.Com vnMath.com Dịch vụ Toán học info@vnmath.com Sách Đại số Giải tích Hình học Các loại khác Chuyên đề Toán Luyện thi Đại học Bồi dưỡng HSG Đề thi Đáp án Đại học Cao học Thi lớp 10 Olympic Giáo án các môn ... nghịch và vì vậy I = (A + B)A −1 (I − C) = A −1 (A + B)(I − C), 4 Dịch Vụ Toán Học Đáp án Đề thi Olympic Toán Sinh viên năm 2010 Đại số và Giải...

Ngày tải lên: 15/12/2013, 01:16

... Show that there exists k ≥ 1 such that F k = 0. 1 PROBLEMS AND SOLUTIONS First day — July 29, 1994 Problem 1. (13 points) a) Let A be a n ì n, n 2, symmetric, invertible matrix with real positive ... at least one is negative. Hence we have at least two non-zero elements in every column of A −1 . This proves part a). For part b) all b ij are zero except b 1,1 = 2, b n,n = (−1) n , b i,i+1 =...

Ngày tải lên: 21/01/2014, 21:20

... − 1 3 . From the hypotheses we have  1 0  1 x f(t)dtdx ≥  1 0 1 − x 2 2 dx or  1 0 tf(t)dt ≥ 1 3 . This completes the proof. Problem 3. (15 points) Let f be twice continuously differentiable on (0, ... |z| = 1. International Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1995 5 if |x − 1| < δ p . From the symmetry we have that (∗) also holds when |y −1| <...

Ngày tải lên: 21/01/2014, 21:20

... consider k ≥ 2. For any m we have (2) cosh θ = cosh ((m + 1)θ − mθ) = = cosh (m + 1)θ.cosh mθ − sinh (m + 1)θ .sinh mθ = cosh (m + 1)θ.cosh mθ −  cosh 2 (m + 1)θ − 1. √ cosh 2 mθ − 1 Set cosh kθ = a, ... finite number of generators all of them cannot be achieved. 1 PROBLEMS AND SOLUTIONS First day — August 2, 1996 Problem 1. (10 points) Let for j = 0, . . . , n, a j = a 0 + jd, where a...

Ngày tải lên: 21/01/2014, 21:20

... have 1 + 1 n k − 1 < θ k 1 . Then we get α (1 + 1 n 1 ) (1 + 1 n 2 ) . . . = θ k 1 (1 + 1 n k ) (1 + 1 n k +1 ) . . . ≥ θ k 1 (1 + 1 n k ) (1 + 1 n 2 k ) . . . = θ k 1 1 + 1 n k 1 > 1 – a ... θ k 1 then θ k ≤ 1, θ k +1 < 1 and hence {θ k } does not converge to 1. Now observe that for M > 1, (2)  1 + 1 M  1 + 1 M 2  1 +...

Ngày tải lên: 21/01/2014, 21:20

... 4c 2 x 2 + O(x 3 ), 2f(x)f  (x) = 4c 2 x 2 + O(x 3 ) and 2( f  (x)) 2  f(x) = 2( 4c 2 x 2 + O(x 3 ))|x|  c + O(x). g is bounded because 2( f  (x)) 2  f(x) |x| 3 −→ x→0 8c 5 /2 = 0 and f  (x) 2 − ... C 2 (R)? Solution. Let c = 1 2 f  (0). We have g = (f  ) 2 − 2ff  2( f  ) 2 √ f , where f(x) = cx 2 + O(x 3 ), f  (x) = 2cx + O(x 2 ), f  (x) =...

Ngày tải lên: 21/01/2014, 21:20

... contains only powers of ¯y = yK, i.e., S 4 /K is cyclic. It is easy to see that this factor-group is not comutative (something more this group is not isomorphic to S 3 ). 3) n = 5 a) If x = (12), then for ... that x is uniquely determined by the triple u, v, w; since the set of such triples is countable, this will finish the proof. To prove the claim, suppose, that from some x  ∈ M we arriv...

Ngày tải lên: 26/01/2014, 16:20

... opposite inequality holds ∀m  1. This contradiction shows that g is a constant, i.e. f(x) = Cx, C > 0. Conversely, it is easy to check that the functions of this type verify the conditions ... contains at most one marked point, delete it. This decreases n + k by 1 and the number of the marked points by at most 1, so the condition remains true. Repeat this step until each row and column co...

Ngày tải lên: 26/01/2014, 16:20

... e, this yields e + 2ef = 0, and similarly f + 2ef = 0, so that f = −2ef = e, hence e = f = g by symmetry. Hence, finaly, 3e = e + f + g = 0, i.e. e = f = g = 0. For part (i) just omit some of this. Problem ... c 0 I + CA. It is well-known that the characteristic polynomials of AC and CA are the same; denote this polynomial by f(x). Then the characteristic polynomials of matrices q(e AB ) and q...

Ngày tải lên: 26/01/2014, 16:20

... ·  g(x)  A     g(x)  A+1   = B A + 1 . By l’Hospital’s rule this implies lim x→∞ f(x) g(x) = lim x→∞ f(x) ·  g(x)  A  g(x)  A+1 = B A + 1 . 4 8 th IMC 2001 July 19 - July 25 Prague, Czech Republic Second ... 4/x 2   . It is a matter of simple manipulation to prove that 2f(x) > x for all x ∈ (0, 2), this implies that the sequence (2 n a n ) is strictly 1 increasing. The i...

Ngày tải lên: 26/01/2014, 16:20

... Let x n = 2 n n  1  n 1 0  + 1  n 1 1  + ···+ 1  n 1 n 1   ; then x n +1 = 2 n +1 n + 1 n  k=0 1  n k  = 2 n n + 1  1 + n 1  k=0  1  n k  + 1  n k +1   + 1  = = 2 n n + 1 n 1  k=0 n−k n + k +1 n  n 1 k  + 2 n +1 n ... 1  = = 2 n n + 1 n 1  k=0 n−k n + k +1 n  n 1 k  + 2 n +1 n + 1 = 2 n n n 1  k=0 1  n 1 k  + 2 n +1...

Ngày tải lên: 26/01/2014, 16:20

... ĐỀ THI OLYMPIC NĂM HỌC 2010 – 2011 MÔN: SINH HỌC - LỚP 8 Thời gian làm bài 120 phút Câu 1. a) Bạch cầu đã tạo ra những hàng

Ngày tải lên: 25/06/2015, 06:00