Đề giao lưu học sinh giỏi cấp Huyện song ngữ môn Toán 7 năm học 2022-2023

Nội dung text: Đề giao lưu học sinh giỏi cấp Huyện song ngữ môn Toán 7 năm học 2022-2023

1. PHÒNG ĐỀ GIAO LƯU HỌC SINH GIỎI CẤP HUYỆN ĐỀ ĐỀ XUẤT NĂM HỌC Đ12 (Đề thi có 02 trang) MÔN: TOÁN LỚP 7 Ngày thi: ./ /2023 Thời gian làm bài 150 phút, không kể thời gian giao đề A- PART I. TESTS. Question 4. Find the unit digit of 172000 ? A. 9 B. 3 C. 7 D. 1 2024 Question 2. Know that the pair of numbers satisfy x 1 y 1 0.Then x2024 y2024 equal to: A. 0 B. 1 C. 2 D. 3 Question 3. The greatest general divisor of 14n + 3 and 21n + 4 is: A. 1 B. 7 C. -1 D. 6 Question 4: The graph of the function y = 3x + 2 passes through A. (3;3) B. (-1; 0) C . (0;0) D. (2;6) Question5. Given the table of initial statistics as follows: 6 7 8 9 8 9 7 6 5 4 7 8 8 9 8 9 7 8 9 4 The value with the highest frequency in the original statistics table above is: A. 9 B. 8 C.7 D. 6 Question 6. Consider a triangle with perimeter 13cm. Knowing the lengths of the 3 altitudes are 2cm, 3cm, 4cm respectively. The length of the largest side of the triangle is: A.6 B. 5 C. 4 D. 3 Question 7. A vertical prism has a base of a rhombus of side 6cm and the surrounding area of the prism is 192 (cm2). Then the height of the prism is equal to: A. 8cm B. 12cm C. 16cm D. 48cm Question 8. Let , Â = 400. The bisector of angle B and angle C intersect at I. The measure of angle BIC is: A. 1500 B. 1400 C. 1300 D. 1100 Question 9. The three friends picked up 210 chestnuts and decided to divide them according to the age ratio. For every friend An gets 4 seeds, friend Ha gets 3 seeds and after friend An gets 6 seeds, friend Lan gets 7 seeds. So how many chestnuts does Ha get? A. 72 B. 54 C. 55 D. 80
2. Question 10. Among 100 people Who are being interviewed, 71 like music and 66 like sports. How many people like both music and sports? A. 29 B. 34 C. 37 D. 27 B- PART II. ESSAY Lesson 1 1. A road repair worker unit intends to divide the number of meters of road to be repaired by 3 groups: Group 1, Group 2, Group 3 respectively in the ratio 4: 5 : 6. But then, because the number of people changed. Therefore, the unit has re-divided the number of meters of road to be repaired for Group 1, Group 2, and Group 3 respectively according to the ratio of 3: 4 : 5. Therefore, one team made 20m less than planned. Calculate the number of unit line meters divided back for each nest.. 2.Find the minimum value of the expression:C = |x 2017| 2018 . |x 2017| 2019 Lesson 2 Let ABC be an isosceles triangle with base BC. Let M and N be the midpoints of AB and AC respectively. Draw NH perpendicular to CM at H. Draw HE perpendicular to AB at E. Draw AK perpendicular to CM at K. Draw AQ perpendicular to HN at Q. 1. Prove that AK = HC = AQ. Calculate the measure of angle BKA. 2. Prove that triangle ABH is isosceles and HM is the bisector of angle BHE. Lesson 3 2 1. Find a natural number ab such that ab (a b)3 2. Find the integers x, y satisfying the equality: (x y)2 2(xy y2 4y) xy y2 4y. Lesson 4 1.Let x, y, z be 3 arbitrary real numbers satisfying : x + y + z = 0 và 1 x 1 , 1 y 1 , 1 z 1. Prove that the polynomial x 2 y 4 z 6 has a value not greater than 2. 2.Given 20 non-zero integers: a1, a2, a3, , a20 have the following properties: *a1 is a positive number. * The sum of any three consecutively written numbers is a positive number. * The sum of those 20 numbers is negative. Prove that: a1.a14 + a14a12 < a1.a12 ======THE END======
3. PHÒNG GD&ĐT VĨNH BẢO ĐÁP ÁN THANG ĐIỂM ĐỀ ĐỀ XUẤT GIAO LƯU HỌC SINH GIỎI CẤP HUYỆN (Đề thi có 02 trang) NĂM HỌC 2022-2023 MÔN: TOÁN LỚP 7 Ngày thi: ./ /2023 Thời gian làm bài 150 phút, không kể thời gian giao đề A- PART I. TESTS. (100 điểm) Câu 1 2 3 4 5 6 7 8 9 10 Đáp án D C A D B A A D B C Điểm 10,0 10,0 10,0 10,0 10,0 10,0 10,0 10,0 10,0 10,0 B- PART II. ESSAY. (200,0 điểm) Leson Solution Marks Leson1 (50,0) Let's call the total number of meters of unit roads that workers have to fix as M (M > 0). Let x1, y1, z1 be the number of meters of the road to be repaired by Group 1, Group 2, and Group 3 respectively as originally 5,0 planned. Let x2, y2, z2 be the number of meters of the road to be repaired by Group 1, Group 2, and Group 3 respectively when redistributed. According to the article we have x1 + y1 + z1 = x2 + y2 + z2 = M (1). 1 5,0 x1 y1 z1 x2 y2 z2 (30Marks) (2); (3) 4 5 6 3 4 5 From (1), (2), (3), applying the property of the sequence of 5,0 equal ratios, we have: x y z x y z M 1 1 1 1 1 1 4 5 6 15 15 5,0 x y z x y z M 2 2 2 2 2 2 3 4 5 12 12 4M M 2M M M 5M x , y ,z ; x , y ,z 5,0 1 15 1 3 1 5 2 4 2 3 2 12
4. According to the output we have:x1 – x2 = 20. 4M M 20 M = 1200. 15 4 So, the number of meters of unit roads that have been divided 5,0 among Group 1, Group 2, and Group 3 are 300m, 400m, and 500m, respectively. x 2017 2019 1 a) C = | x 2017 | 2018 = = | x 2017 | 2019 | x 2017 | 2019 5,0 1 1 | x 2017 | 2019 The expression C reaches its minimum value when 1 | x 2017 | 2019 has the smallest value 5,0 (20Marks) Because | x 2017 | ≥ 0 so | x 2017 | 2019 ≥ 2019. 2018 The equality holds when x = 2017 C = . 2019 5,0 2018 So the minimum value of C is when x = 2017. 5,0 2019 Leson2 50,0 1 (50Marks) 10
5. 1. Considering two right triangles MKA and NHC, there are: 20Marks 1 1 AM = CN ( AB AC ) 2 2 5,0 MAK NHC ( sub with AMC) MKA = NCH (hypotenuse, acute angle) KA = HC (two corresponding edges) (1) - prove: AQN = CHN (hypotenuse, acute angle) 5,0 AQ = CH (two corresponding edges) (2) From (1) and (2) KA = HC = AQ. 5,0 - Prove AKH = AQH (hypotenuse, right angle side) AHC and BKA have : AC = AB; HC = AK BAK = HCA (sub with AMC). 5,0 Deduce : AHC = BKA (c.g.c)  BKA = AHC = 1350. BKA and BKH have AK = KH (because quadrilateral AKHQ is a square). BKH = 3600 – AKH – AKB = 1350 = AKB. 10,0 BK is the common edge. BKA = BKH (c.g.c). AB = BH (two corresponding edges) 5,0 2 AHB is an isosceles triangle at B. 30Marks We have : MHE = HCA (2 isotopic angles, EH//AC because they are perpendicular to AB). HCA = KAB ( AHC = BKA) 10,0 KAB = KHB ( BKA = BKH) MHE = KHB HM is the bisector of the angle BHE. Leson3 50,0 2 We have : (a + b)3 = ab is a perfect square, so a + b is a perfect square. Put: a + b = x2 (x N * ) 2 ab (a b)3 = x6 5,0 1 (20Marks) => x3 = ab 8 => 8 2 x = 3; 4 becase x N *
6. 2 -If x = 3 => ab (a b)3 = 36 = 729 = 272 = (2 + 7)3 => x = 3 5,0 (accept) 2 -If x = 4 => ab (a b)3 = 46 = 4096 = 642 (6 + 4)3 = 1000 => x = 4 (no sense) 10,0 So: ab = 27 (x y)2 2(xy y2 4y) xy y2 4y (1) (x y)2 2(xy y2 4y) xy y2 4y (1) 5,0 (x y)2 2(xy y2 4y) 0 For all x, y, we have , xy y2 4y 0 From (1) (x y)2 2(y2 xy 4y) 0 5,0 2 (x y)2 2(xy y2 4y) xy y2 4y (30Marks) From (1) 5,0 (x y)2 xy y2 4y 0 5,0 (x y)2 0 2 2 5,0 2 (Because (x y) 0, y xy 4y 0) y xy 4y 0 x = y = 0 or x = y = 2. So (x; y) is: (0; 0), (2; 2). 5,0 Lesson 4 50,0 +) Of the three numbers x, y, and z, at least two have the same sign. Suppose x; y 0 10,0 1 => z = - x - y 0 (30Marks) +) Because 1 x 1 , 1 y 1 , 1 z 1 = > x2 y4 z6 x y z 10,0 => x 2 y 4 z 6 x y z => x 2 y 4 z 6 2z 10,0 +) 1 z 1 and z 0 => x 2 y 4 z 6 2 So: x 2 y 4 z 6 2
7. We have a + (a + a + a ) + + (a + a + a ) + a + (a + a + 1 2 3 4 11 12 13 14 15 16 5,0 a17) + (a18 + a19 + a20) 0 ; a2 + a3 + a4 > 0 ; ; a11 + a12 + a13 > 0 ; a15 + a16 + a17 > 0 ; 2 a18 + a19 + a20 > 0 => a14 < 0. (20Marks) So : (a + a + a ) + + (a + a + a ) + (a + a ) + (a + 1 2 3 10 11 12 13 14 15 5,0 a16 + a17) + (a18 + a19 + a20) a13 + a14 < 0. On the other hand,a12 + a13 + a14 > 0 => a12 > 0. 5,0 From the conditions : a > 0 ; a > 0 ; a a .a + a a 1 12 14 1 14 14 12 5,0 < a1.a12 (đpcm). Lưu ý khi chấm bài: - Trên đây chỉ là sơ lược các bước giải, lời giải của học sinh cần lập luận chặt chẽ, hợp logic. Nếu học sinh trình bày cách làm khác mà đúng thì cho điểm các phần theo thang điểm tương ứng. - Nếu thí sinh vẽ sai hình thì không chấm điểm.