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 12 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 12345678910111213141516171819202122232425 Vaqtingiz tugadi! 10-sinf Matematika olimpiada №3 2021-2022 o'quv yili viloyat bosqichida tushgan savollar. 1 / 25 1. Tenglama ildizlarining o‘rta arifmetik qiymatini toping: A) 2 B) 5 C) 1 D) 3 2 / 25 2. trigonometrik tenglamani yeching. A) π/6+πk; k∈Z B) π/12+2πk; k∈Z C) π/12+πk/2; k∈Z D) π/12+πk; k∈Z 3 / 25 3. Agar ni toping. A) 0.96 B) 0.8 C) 0.81 D) 0.9 4 / 25 4. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 5 B) 3 C) 8 D) 2 5 / 25 5. 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) 8 B) 7 C) 6 D) 5 6 / 25 6. Agar bo‘lsa, |xyz| ni toping. A) 1000 B) 100 C) 800 D) 500 7 / 25 7. Tenglama nechta yechimga ega? A) 2 B) 4 C) 3 D) 5 8 / 25 8. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 1 B) 0 C) -1 D) 3 9 / 25 9. 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) 11 C) 8 D) 10 10 / 25 10. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) √2 B) √3 C) 1 D) 2√2 11 / 25 11. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 15 B) 10 C) 12 D) 9 12 / 25 12. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) sin18° B) 2cos² 9° C) tg9° D) 1 13 / 25 13. 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) 8 B) 10 C) 15 D) 9 14 / 25 14. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 16 B) 12 C) 8 D) 0 15 / 25 15. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x ≤ 3 B) x<0 C) 3 ≤ x D) x>3 16 / 25 16. 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) 120° B) 135° C) 90° D) 150° 17 / 25 17. x, y sonlari (x² + 1)(y² + 1) + 2(x − y)(1 − xy) = 4(1 + xy) tenglikni qanoatlantiradi |1 + x| ∙ |1 − y| ni toping. A) 2 B) 1 C) 9 D) 3 18 / 25 18. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 1 B) -2 C) 0 D) -1 19 / 25 19. Agar A) π/2 B) π C) 0 D) 3π/2 20 / 25 20. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -2 B) 3 C) -3 D) 2 21 / 25 21. Agar ?1 = 6? − 6 va у2//?1 hamda у2 to‘g‘ri chiziq ?(6; 6) nuqtadan o‘tsa, ?2 ni toping. A) −6? + 6 B) 6? − 30 C) −6? + 42 D) 6? − 24 22 / 25 22. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 35 B) 2√21 C) 4√21 D) √21 23 / 25 23. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 10? + 2 B) 4? + 2 C) 8? + 2 D) 6? + 2 24 / 25 24. 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√6/24 B) 2/5 C) 5√3/12 D) 5√2/5 25 / 25 25. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 5 B) 6 C) 4 D) 7 0% Testni qayta ishga tushiring Tomonidan Wordpress Quiz plugin Author: InfoMaster Foydali bo'lsa mamnunmiz
