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. ABCD trapetsiyaning balandligi 10 ga teng va asoslari AD:BC=7:3 shartni qanoattlantiradi. Trapetsiyaning diagonallari kesishgan nuqtadan katta asosgacha bo‘lgan masofani toping. A) 7 B) 10 C) 9 D) 8 2 / 25 2. Uchburchakning ikki tomoni 12 va 13 sm ga, ular orasidagi burchak kosinusi 5/13 ga teng bo`lsa, uning yuzini toping. A) 71 B) 60 C) 72 D) 70 3 / 25 3. bo‘lsa ? ning qiymatini toping. A) 37 B) 25 C) 12 D) 40 4 / 25 4. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) x+25 B) 20x+48 C) 0 D) 20x+50 5 / 25 5. Agar A) π/2 B) 3π/2 C) π D) 0 6 / 25 6. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 16 B) 15 C) 17 D) 14 7 / 25 7. 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) 7 B) 5 C) 8 D) 6 8 / 25 8. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 1 B) -1 C) 3 D) 0 9 / 25 9. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 2cos² 9° B) sin18° C) 1 D) tg9° 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√2 B) √3 C) √2 D) 1 11 / 25 11. Agar bo‘lsa, |xyz| ni toping. A) 800 B) 1000 C) 100 D) 500 12 / 25 12. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 6 B) 4 C) 7 D) 5 13 / 25 13. 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) 12 D) 10 14 / 25 14. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -3 B) 2 C) 3 D) -2 15 / 25 15. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x<0 B) x ≤ 3 C) 3 ≤ x D) x>3 16 / 25 16. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 35 B) √21 C) 2√21 D) 4√21 17 / 25 17. Agar ?1 = 6? − 6 va у2//?1 hamda у2 to‘g‘ri chiziq ?(6; 6) nuqtadan o‘tsa, ?2 ni toping. A) −6? + 42 B) −6? + 6 C) 6? − 24 D) 6? − 30 18 / 25 18. Tenglamalar sistemasi nechta yechimga ega? A) 10 B) 12 C) 8 D) 6 19 / 25 19. 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) 135° B) 150° C) 120° D) 90° 20 / 25 20. 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 21 / 25 21. x, y sonlari (x² + 1)(y² + 1) + 2(x − y)(1 − xy) = 4(1 + xy) tenglikni qanoatlantiradi |1 + x| ∙ |1 − y| ni toping. A) 1 B) 3 C) 2 D) 9 22 / 25 22. 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) 8 C) 9 D) 15 23 / 25 23. Agar ni toping. A) 0.96 B) 0.81 C) 0.9 D) 0.8 24 / 25 24. 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) 9 D) 11 25 / 25 25. Tenglama nechta yechimga ega? A) 4 B) 2 C) 3 D) 5 0% Testni qayta ishga tushiring Tomonidan Wordpress Quiz plugin Author: InfoMaster Foydali bo'lsa mamnunmiz
