10-sinf Matematika olimpiada №6 Yanvar 2, 2024Avgust 19, 2024 da chop etilgan InfoMaster tomonidan 10-sinf Matematika olimpiada №6 ga 1 fikr bildirilgan 0 123456789101112131415161718192021222324252627282930 Vaqtingiz tugadi! Yaratilgan sanasi Avgust 19, 2024 tomonidan InfoMaster 10-sinf Matematika olimpiada №6 2023-yil 21-oktyabr: 2023-2024 o'quv yili tuman bosqichida tushgan savollar! 1 / 30 Agarbo’lsa, quyidagilardan qaysi biri to’g’ri? A) 2𝑏 = 𝑎 − 2 B) 𝑏 = 2𝑎 + 2 C) 𝑏 = 𝑎 + 2 D) 𝑎 = 2𝑏 − 2 2 / 30 tgα = 3 bo‘lsa, ifodaning qiymatini toping. A) 5 B) 3 C) 1.5 D) 1 3 / 30 Soddalashtiring: A) √7 B) √2 C) 1 D) 2 4 / 30 Haqiqiy 𝑥, 𝑦 sonlari uchun 𝑥 + 𝑦 = 4 va 𝑥𝑦 = −2 boʻlsa, ni hisoblang. Tekshirish 5 / 30 (𝑥 + 1) ∙ |𝑥 + 5| = 3(𝑥² − 25) tenglamaning haqiqiy ildizlari sonini toping. A) 1 B) 3 C) ∅ D) 2 6 / 30 Hisoblang: A) 1 B) 0.75 C) 3 D) 1.5 7 / 30 𝑥, 𝑦 −natural sonlar 𝑥³ − 𝑥²𝑦 + 𝑥𝑦² − 𝑦3 = 2023 tenglikni qanoatlantiradi. 𝑥 − 2𝑦 ayirmani toping. Tekshirish 8 / 30 201200+2023023 sonining oxirgi ikki raqami yig‘indisini toping. Tekshirish 9 / 30 Hisoblang: A) √3 B) 1 C) 1/2 D) 1/√3 10 / 30 Quyidagilar orasidan juft funksiyalarni ko‘rsating: A) 1,2,4 B) 1,3,4 C) 2,3,5 D) 2,4,5 11 / 30 𝐴𝐵𝐶𝐷𝐸 −qavariq beshburchak. 𝐴𝐹𝑁, 𝐴𝐵𝑁, 𝐵𝑁𝐶, 𝐶𝐾𝐷, 𝐷𝐾𝐸, 𝐹𝐾𝐸 uchburchak yuzlari mosh ravishda 2,3,9,5,4 va 8 ga teng bo‘lsa, 𝐹𝑁𝐶𝐾 to‘rtburchak yuzini toping. Tekshirish 12 / 30 x1 , x2 , x3 , x4 , x5 , x6 lar {-1,-2,-3,4,5,6} to’plamdan olingan turli sonlar. 3·(x1 + x2)+2·(x3 + x4) + (x5 + x6) yig’indining qabul qilishi mumkin bo’lgan eng katta qiymatini toping. Tekshirish 13 / 30 A punktdan B punktga qarab mototsiklchi, shu vaqtning o‘zida B punktdan A punktga qarab velosipedchi yo‘lga chiqdi. Ular B punktdan 4 km masofada uchrashdi, mototsiklchi B punktga yetib kelganda esa velosipedchi A punktdan 15 km masofada bo‘ldi. A va B punktlar orasidagi masofani toping. A) 20 km B) 27 km C) 30 km D) 24 km 14 / 30 𝑛 −natural son va 𝐸𝐾𝑈𝐾(9; 𝑛) + 𝐸𝐾𝑈𝐵(3; 𝑛) = 129 bo‘lsa, 𝐸𝐾𝑈𝐾(𝑛; 9) ni toping. Tekshirish 15 / 30 𝐴𝐵𝐶𝐷 trapetsiyada 𝐵𝐶,𝐴𝐷 − asoslar va 𝐵𝐶: 𝐴𝐷 = 2: 5. Trapetsiya yuzi 56. ABC uchburchak yuzini toping. A) 12 B) 18 C) 16 D) 14 16 / 30 ifodani soddalashtiring. A) 1 B) tg2α C) 2 D) 2tg2α 17 / 30 Soddalashtiring: A) B B) D C) C D) A 18 / 30 𝑎, 𝑏 va 𝑐 sonlari uchun 𝑎 ∶ 𝑏 ∶ 𝑐 = 1: 2: 3 bo‘lsa, (𝑎 + 𝑏): (𝑏 + 𝑐): (𝑐 + 𝑎) ni toping. A) 2: 3 ∶ 4 B) 3 ∶ 5 ∶ 6 C) 3 ∶ 5 ∶ 4 D) 2: 6 ∶ 5 19 / 30 Agar p>5>q>3 bo'lsa, |p+q-7|-|2-q|+|q-p-6| ifodani soddalashtiring. A) 2𝑝 − 𝑞 + 1 B) 2𝑝 + 𝑞 − 15 C) 3𝑞 − 15 D) 2𝑝 − 𝑞 − 11 20 / 30 𝐴𝐵𝐶 − tengyonli to‘g‘ri burchakli uchburchak. Agar 𝐴𝐷 = √2, 𝐵𝐷 = 1, 𝐶𝐷 = 2 bo‘lsa, 𝐴𝐷𝐶 uchburchak yuzini toping. Tekshirish 21 / 30 Agar tengsizlikni yeching. A) (1,5; 4) B) (0; 1,5) C) (0; 3,5) D) (1,5; 3,5) 22 / 30 𝐴𝐵𝐶𝐷 −kvadrat, ∠𝐴𝐸𝐵 = 90° va 𝐴𝐸 = 4 bo‘lsa, 𝐴𝐸𝐷 uchburchak yuzini toping. Tekshirish 23 / 30 𝑥, 𝑦, 𝑧 −haqiqiy sonlar 𝑥²𝑦 + 𝑦²𝑧 + 𝑧²𝑥 = 0 tenglikni qanoatlantirsa, ifodaning qiymatini toping. Tekshirish 24 / 30 Arifmetik progressiyada 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 = 18 va 𝑎²4 − 𝑎²1 = 54 bo’lsa, 𝑎7 ni toping. A) 12.5 B) 15 C) 14 D) 13.5 25 / 30 AB =3, AC =4, BM =MN =CN=2 bo'lsa, MA²+NA² ni toping. A) 9 B) 12 C) 10 D) 8 26 / 30 Tengsizliklar sistemasini yeching: A) A B) D C) B D) C 27 / 30 Agar bo‘lsa, 𝑥 + 𝑦 yig‘indini toping. A) -1 B) -3 C) 0 D) 3 28 / 30 1 + 42 + 43 + ⋯ + 4100 yig’indini 10 ga bo‘lgandagi qoldiq nimaga teng? Tekshirish 29 / 30 𝑦 < 0 bo‘lsa, ifodani soddalashtiring. A) A B) D C) B D) C 30 / 30 𝐴𝐵𝐶 uchburchakda: ∠A-10°=∠B+40°=∠C-36° bo'lsa, ∠B ni toping. A) 22° B) 24° C) 15° D) 18° 0% tomonidan Wordpress Quiz plugin Matematika olimpiada, Tuman 2023-2024 o'quv yili
