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 53 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. Agar ikkita teng tomonli uchburchakning tomonlari uchun munosabat o‘rinli bo‘lsa, katta uchburchak yuzini kichik uchburchak yuziga nisbatini toping. A) 9 B) 7 C) 6 D) 8 2 / 25 2. Uchburchakning tomonlari 7 va 11 ga, uchinchi tomonining medianasi √21 ga teng. Uchburchakning uchunchi tomonini toping. A) 14 B) 12 C) 16 D) 15 3 / 25 3. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 2√21 B) 35 C) 4√21 D) √21 4 / 25 4. 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) 72 C) 144 D) 36 5 / 25 5. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) tg9° B) sin18° C) 1 D) 2cos² 9° 6 / 25 6. Agar ni toping. A) 0.8 B) 0.81 C) 0.96 D) 0.9 7 / 25 7. Tenglama nechta yechimga ega? A) 3 B) 2 C) 4 D) 5 8 / 25 8. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -24 B) -26 C) -25 D) -27 9 / 25 9. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 9 B) 15 C) 10 D) 12 10 / 25 10. bo‘lsa ? ning qiymatini toping. A) 12 B) 40 C) 37 D) 25 11 / 25 11. Agar A) π B) 0 C) π/2 D) 3π/2 12 / 25 12. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 12 B) 0 C) 16 D) 8 13 / 25 13. 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) 2/5 B) 5√6/24 C) 5√3/12 D) 5√2/5 14 / 25 14. Agar bo‘lsa, |xyz| ni toping. A) 800 B) 100 C) 500 D) 1000 15 / 25 15. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) -1 B) 1 C) 0 D) 3 16 / 25 16. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 4? + 2 B) 8? + 2 C) 10? + 2 D) 6? + 2 17 / 25 17. 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) 5 B) 8 C) 6 D) 7 18 / 25 18. Tenglamalar sistemasi nechta yechimga ega? A) 10 B) 8 C) 12 D) 6 19 / 25 19. 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) 10 B) 8 C) 9 D) 11 20 / 25 20. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) 1 B) √2 C) √3 D) 2√2 21 / 25 21. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -3 B) 3 C) -2 D) 2 22 / 25 22. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 6 B) 7 C) 4 D) 5 23 / 25 23. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x ≤ 3 B) 3 ≤ x C) x<0 D) x>3 24 / 25 24. 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) 90° B) 150° C) 135° D) 120° 25 / 25 25. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 8 B) 5 C) 3 D) 2 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
