chuyen de bai tap toan hoc sinh gioi lop 4

... 121816 986 ìì ìì b) 9 ì 4 ì 67 + 6 ì 13 + 12 ì 3 ì 20 Cõu 3. Tỡm x bit a) 9,7 ì x < 9,718 b) 208,1 – x = 32 ,4 + 56 c) 535 35 ì x = 515 Cõu 4. Một đội công nhân có 24 người dự định làm ... ì 9 900 ì 0,1 9) c) Tớnh bng 2 cỏch: (27,8 + 16 ,4) ì 5 Cõu 2. Chuyển các phân số sau về số thập phân: a) 20 12 b) 4 1 c) 25 24 d) 125 25 Câu 3. Đầu xuân kỷ sửu ba bạn An, Bình, ... CHẤT LƯỢNG HỌC SINH GIỎI LỚP 5 ĐỢT 1 - MÔN : TOÁN Tháng 11 Năm học 2008 - 2009 Thời gian 60 phút ( không kể thời gian giao ) Cõu 1. Tớnh a) 2 3 2 ì 5 4 1 b) 3 2 1 : 2 4 1 c) 5 3 ...

... thức a) A = 747 4 −−+ b) B = 521 045 21 04 +−+++ c) C = 5321 541 54 −−−++ Bài 20: Tính giá trị biểu thức P = 12 341 27 24 −−++−++ xxxx Với 2 1 ≤ x ≤ 5. Bài 21: Chứng minh rằng: P = 26 48 13532 + +−+ ... Giải phương trình: x 4 + 2x 3 + 5x 2 + 4x - 5 = 0 Bài 39: Tìm nghiệm nguyên của phương trình: 4x 4 + 8x 2 y + 3y 2 - 4y - 15 = 0. Bài 40 : x 2 + 9x + 20 = 2 103 + x Bài 41 : x 2 + 3x + ... HSG 1998) Bài 29: 3 53 14 5 = −+ − −− x x x Bài 30: x 4 - 4 3 x -5 = 0 ( Đề thi HSG 2000) Bài 31: 05 2 4 2 4 =− − + x x x ( Đề thi HSG V2 2003) Bài 32: a) x 4 - 4x 3 - 19x 2 + 106x - 120...

... B. 27 cm 2 C. 90 cm 2 D. 72 cm 2 D 42 Một gói đường mua vào 20000 đồng. Hỏi phải bán với giá bao nhiêu để lãi 20% giá bán? 25000 đồng 43 Cho phân số 40 4 128 . Phải bớt ở tử số để thêm vào ... 5 2 ? A. 12 B. 24 C. 36 D. 30 B 44 Nếu giảm độ dài đáy của một tam giác đi 3 1 thì diện tích tam giác đó giảm đi 127 cm 2 . Vậy diện tích ban đầu của tam giác đó là bao nhiêu? A. 348 cm 2 B. ... 100 B. Chia cho 100 C. Nhân với D. Chia cho A 40 Tìm một số chia cho 3 dư 2, chia cho 4 dư 3 và chia hết cho 5. Số đó là: A. 35 B. 11 C. 15 D. 45 A 41 A B C D E F 22 Tỉnh Hải Dương có một thầy...

... 0, 34 mol = 0,17.2 = 0, 34 mol  n hhA = 9 ,40 8 : 22 ,4 = 0 ,42 mol n hhA = 9 ,40 8 : 22 ,4 = 0 ,42 mol  Goïi x , y , z là số mol C Gọi x , y , z laø soá mol C 2 2 H H 2 2 , C , C 2 2 H H 4 4 ... C C 2 2 H H 2 2 + 2Br + 2Br 2 2   C C 2 2 H H 2 2 Br Br 4 4 C C 2 2 H H 4 4 + Br + Br 2 2   C C 2 2 H H 4 4 Br Br 2 2  nC nC 6 6 H H 6 6 = 1,56 : 78 = 0,02 mol => ... = 0,5 mol ⇒ Goïi a mol C Goïi a mol C 2 2 H H 4 4 đã phản ứng đã phản ứng Theo PT : n C Theo PT : n C 2 2 H H 6 6 = n C = n C 2 2 H H 4 4 = a mol = a mol t 0 Ni Bài giải Bài...

... thu được sắt và khí CO 2 . -Nếu cho lượng sắt ở trên vào dd HNO 3 đặc , nóng , dư thu được 13 ,44 lít khí NO 2 ở đktc và dd chứa Fe(NO 3 ) 3 . -Nếu cho khí CO 2 hấp thụ hết vào dd Ca(OH) 2 ... = a + (5a/68)32 – (6a/68)18 = 120a /68 (g) ⇒ n NO = m NO : 30 = 4a/68 (mol) Ta coù : m O = m O(NO) + m O(H2O) = (4a/68).16 + (6a/68).16 = 160a/68 Maø : m O(O2) = 5a/68 .32 = 160a/68 ... A là hợp chất của N và H . Gọi công thức hóa học của A là N x H y . ⇒ Ta có tỉ lệ : x : y = (4a/68): (12a/68) x : y = 1 : 3 ⇒ A laø (NH 3 ) n . Maø A chứa 1 nguyên tử N ⇒ A là NH 3 ...

... khoảng cách từ điểm O đến AB bằng a và SAO = 30 0 , SAB = 60 0 . tính độ dài đường sinh của hình nón theo a. Bài 4: Cho khối nón tròn xoay có đường cao h = 20cm, bán kính đáy R = 25cm. một mặt phẳng ... thiết diện đó. Bài 5: Một mặt phẳng ( ) α đi qua hai đường sinh của hình nón cắt mặt đáy hình nón theo một dây cung có độ dài gấp 4 lần đường cao của hình nón. Tính góc ϕ giữa mặt phẳng ( ... diện tích xung quanh của hình nón này theo a và α . Bài 12: Một hình nón có đường sinh là l và góc giữa đường sinh và đáy là α . a) Tình diện tích xung quanh và thể tích của khối nón. b) Gọi...

... were lost forever - some books were ( 14) ……… recovered . We can not even know (15) ………. exactly the library contained . III.Read the text below and decide which answer ( A , B , C or D ) best ... consideration 10. A. escaping B. hiding C. running D. disappearing 11. A. because B. when C. and D. so 12. A. die B. preserve C. survive D. continue 13. A. discuss B. talk C. solve D. decide ... simple : books (4) ……… a very long time to produce , and there were far (5) ………… copies of any given work around . The greatest library of (6) ………. that in Alexandria , had 54, 000 books . In...

... khan C 14 MgBr C 14 MgBr CO 2 COOMgBr COOMgBr H 3 O COOH C 14 COOH HNO 3 H 2 SO 4 C 14 COOH NO 2 HOH + (1) + (2) + + (3) + + (4) C 14 COOH NO 2 Na 2 CO 3 C 14 NO 2 COONa HOH CO 2 C 14 NO 2 COONa CaO C 14 NO 2 Na 2 CO 3 C 14 NO 2 C 14 NH 3 Cl FeCl 3 HOH C 14 NH 3 Cl NaOH C 14 NH 2 NaCl HOH + + + (5) + NaOH ... hữu cơ b, C 14 NH 2 NaNO 2 C 14 N N C 14 I N 2 KCl - C 14 N N - + + 2 HCl + + NaCl + 2HOH + KI + + (9) (10) Cl + Cl c, C 14 N N C 14 C N N 2 CuCl H + C 14 COOH NH 4 C 14 C N + + CuCN - + + 2HOH + ... khan C 14 MgBr C 14 MgBr CO 2 COOMgBr COOMgBr H 3 O COOH C 14 COOH HNO 3 H 2 SO 4 C 14 COOH NO 2 HOH + (1) + (2) + + (3) + + (4) C 14 COOH NO 2 Na 2 CO 3 C 14 NO 2 COONa HOH CO 2 C 14 NO 2 COONa CaO C 14 NO 2 Na 2 CO 3 C 14 NO 2 C 14 NH 3 Cl FeCl 3 HOH C 14 NH 3 Cl NaOH C 14 NH 2 NaCl HOH + + + (5) + NaOH rắn + (6) 2 2 + 3 Fe + 7 HCl...

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... HCl, HNO 3 , H 2 SO 4 . d) Na 2 SO 4 , NaCl, NaNO 3 e) HNO 3 , H 2 SO 4 , KCl, KNO 3 , KOH, Ba(OH) 2 . g) K 2 SO 4 , CuSO 4 , FeSO 4 , Fe 2 (SO 4 ) 3 , Al 2 (SO 4 ), MgSO 4 . h) CO 2 , H 2 , ... kim loại Mg và Zn. (B) là dd H 2 SO 4 có nồng độ laø x mol/l. TN1: cho 34, 02g A vaøo 2,8 lít B sinh ra 13 ,44 lít khí H 2 . TN2: cho 34, 02g A vaøo 4, 2 lít B sinh ra 15,68 lít khí H 2 . a. Haõy ... -> SO 3 -> H 2 SO 4 -> CuSO 4 -> Na 2 SO 4 -> NaOH -> Na 2 ZnO 2 . h) P -> P 2 O 5 -> H 3 PO 4 -> NaH 2 PO 4 -> Na 2 HPO 4 –> Na 3 PO 4 -> NaCl -> NaOH. i)...