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 47 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. tengsizlikning eng katta butun yechimini toping. A) 1 B) 4 C) 3 D) 2 2 / 25 2. trigonometrik tenglamani yeching. A) π/12+πk; k∈Z B) π/6+πk; k∈Z C) π/12+2πk; k∈Z D) π/12+πk/2; k∈Z 3 / 25 3. 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) 8 B) 10 C) 11 D) 9 4 / 25 4. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 15 B) 14 C) 17 D) 16 5 / 25 5. Agar A) 0 B) π/2 C) π D) 3π/2 6 / 25 6. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) -2 B) 0 C) -1 D) 1 7 / 25 7. Agar ni toping. A) 0.8 B) 0.96 C) 0.9 D) 0.81 8 / 25 8. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 5 B) 6 C) 7 D) 4 9 / 25 9. bo‘lsa ? ning qiymatini toping. A) 12 B) 37 C) 25 D) 40 10 / 25 10. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 20x+50 B) 20x+48 C) 0 D) x+25 11 / 25 11. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -25 C) -24 D) -26 12 / 25 12. 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 13 / 25 13. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 2√21 B) √21 C) 4√21 D) 35 14 / 25 14. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 10 B) 15 C) 9 D) 12 15 / 25 15. 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√6/24 C) 5√2/5 D) 2/5 16 / 25 16. 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) 8 C) 7 D) 5 17 / 25 17. ABCD to‘rtburchakda АВ = CD = 9 va bu to‘rtburchakka radiusi 4 ga teng aylana ichki chizilgan. ABCD to‘rtburchak yuzini toping. A) 36 B) 72 C) 144 D) 81 18 / 25 18. 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) 2√2 D) √3 19 / 25 19. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 1 B) 0 C) 3 D) -1 20 / 25 20. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 2cos² 9° B) 1 C) tg9° D) sin18° 21 / 25 21. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x<0 B) 3 ≤ x C) x ≤ 3 D) x>3 22 / 25 22. Tenglamalar sistemasi nechta yechimga ega? A) 6 B) 10 C) 12 D) 8 23 / 25 23. Agar bo‘lsa, |xyz| ni toping. A) 1000 B) 100 C) 500 D) 800 24 / 25 24. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) 2 B) -3 C) -2 D) 3 25 / 25 25. 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) 120° C) 135° D) 90° 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
