Uy » Online olimpiada » Matematika olimpiada » 10-sinf Matematika olimpiada №3 Matematika olimpiadaViloyat 2021-2022 o'quv yili 10-sinf Matematika olimpiada №3 InfoMaster Avgust 19, 2024 123 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Savollar 12345678910111213141516171819202122232425 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster 10-sinf Matematika olimpiada №3 2021-2022 o'quv yili viloyat bosqichida tushgan savollar. 1 / 25 1. ko‘phadni standart ko‘rinishda ifodalangandagi koeffitsientlar yig‘indisini toping. 1. 2. 3. 4.1 A) 3 B) 2 C) 1 D) 4 2 / 25 2. tengsizlik yechimi bo`la oladigan tub sonlar nechta? A) 3 B) 2 C) 4 D) 5 3 / 25 3. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 0 B) 16 C) 8 D) 12 4 / 25 4. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 15 B) 16 C) 14 D) 17 5 / 25 5. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 3 B) 8 C) 5 D) 2 6 / 25 6. Tenglama nechta yechimga ega? A) 5 B) 2 C) 3 D) 4 7 / 25 7. ABCD to‘rtburchakda АВ = CD = 9 va bu to‘rtburchakka radiusi 4 ga teng aylana ichki chizilgan. ABCD to‘rtburchak yuzini toping. A) 81 B) 144 C) 36 D) 72 8 / 25 8. ABCD to‘rtburchakda А va В burchaklar- to‘g‘ri, tg∠D = 3/4 va ВС = AD/2 = АВ + 2 bo‘lsa, АС ni toping. A) 10 B) 9 C) 8 D) 15 9 / 25 9. ABC – gipotenuzasi AB bo‘lgan to‘g‘ri burchakli uchburchak. Gipotenuzaning ikki tomon davomida AB to‘g‘ri chiziqda AK = AC va BM = BC shartlar bilan kesmalar ajratilgan. KCM burchakni toping. A) 150° B) 90° C) 120° D) 135° 10 / 25 10. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 2√21 B) 4√21 C) 35 D) √21 11 / 25 11. Agar cos ∠A = 1/5 va sin∠B = 1/2 bo‘lsa, АВС uchburchakning mos ravishda А va В uchlaridan tushirilgan balandliklar nisbatini toping. A) 5√3/12 B) 5√2/5 C) 5√6/24 D) 2/5 12 / 25 12. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -24 C) -26 D) -25 13 / 25 13. Agar ni toping. A) 0.8 B) 0.96 C) 0.9 D) 0.81 14 / 25 14. bo‘lsa ? ning qiymatini toping. A) 12 B) 37 C) 40 D) 25 15 / 25 15. Agar bo‘lsa, |xyz| ni toping. A) 500 B) 1000 C) 800 D) 100 16 / 25 16. Tenglamalar sistemasi nechta yechimga ega? A) 12 B) 8 C) 6 D) 10 17 / 25 17. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) x+25 B) 0 C) 20x+50 D) 20x+48 18 / 25 18. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 1 B) -1 C) -2 D) 0 19 / 25 19. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) tg9° B) 2cos² 9° C) sin18° D) 1 20 / 25 20. Agar A) π/2 B) π C) 3π/2 D) 0 21 / 25 21. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x>3 B) 3 ≤ x C) x ≤ 3 D) x<0 22 / 25 22. Agar ?1 = 6? − 6 va у2//?1 hamda у2 to‘g‘ri chiziq ?(6; 6) nuqtadan o‘tsa, ?2 ni toping. A) 6? − 24 B) 6? − 30 C) −6? + 42 D) −6? + 6 23 / 25 23. Ayirmasi noldan farqli bo‘lgan arifmetik progressiyaning 4-hadidan boshlab 14- hadigacha bo‘lgan hadlar yig‘indisi 77 ga teng. Progressiyaning 7 ga teng bo‘lgan had nomerini toping. A) 9 B) 8 C) 11 D) 10 24 / 25 24. Qavariq to‘rtburchakning diagonallari 3 va 4 ga teng. Agar qarama qarshi tomonlarining o‘rtalarini tutashtirishdan hosil bo‘lgan kesmalar uzunliklari o‘zaro teng bo‘lsa, qavariq to‘rtburchakning yuzini toping. A) 6 B) 7 C) 8 D) 5 25 / 25 25. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -2 B) 3 C) -3 D) 2 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
