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 99 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 yechimlari joylashgan interval uzunligini toping. A) 1,15 B) 1,45 C) 1,35 D) 1,25 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) 60 B) 71 C) 70 D) 72 3 / 25 3. bo‘lsa ? ning qiymatini toping. A) 25 B) 40 C) 12 D) 37 4 / 25 4. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 14 B) 16 C) 15 D) 17 5 / 25 5. 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√2 C) √2 D) √3 6 / 25 6. x, y sonlari (x² + 1)(y² + 1) + 2(x − y)(1 − xy) = 4(1 + xy) tenglikni qanoatlantiradi |1 + x| ∙ |1 − y| ni toping. A) 3 B) 1 C) 9 D) 2 7 / 25 7. Agar A) 0 B) π/2 C) π D) 3π/2 8 / 25 8. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) -1 B) 1 C) 0 D) -2 9 / 25 9. 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 10 / 25 10. Agar ?1 = 6? − 6 va у2//?1 hamda у2 to‘g‘ri chiziq ?(6; 6) nuqtadan o‘tsa, ?2 ni toping. A) 6? − 30 B) 6? − 24 C) −6? + 42 D) −6? + 6 11 / 25 11. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -3 B) -2 C) 2 D) 3 12 / 25 12. 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) 10 C) 11 D) 8 13 / 25 13. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 3 B) 0 C) 1 D) -1 14 / 25 14. Agar bo‘lsa, |xyz| ni toping. A) 800 B) 1000 C) 100 D) 500 15 / 25 15. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 35 B) 4√21 C) 2√21 D) √21 16 / 25 16. Agar ni toping. A) 0.81 B) 0.8 C) 0.96 D) 0.9 17 / 25 17. 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) 2/5 D) 5√6/24 18 / 25 18. 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) 9 B) 10 C) 8 D) 15 19 / 25 19. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 1 B) tg9° C) 2cos² 9° D) sin18° 20 / 25 20. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -24 C) -25 D) -26 21 / 25 21. 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) 9 D) 12 22 / 25 22. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) 3 ≤ x B) x<0 C) x>3 D) x ≤ 3 23 / 25 23. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 5 B) 4 C) 7 D) 6 24 / 25 24. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 16 B) 12 C) 8 D) 0 25 / 25 25. Tenglamalar sistemasi nechta yechimga ega? A) 8 B) 12 C) 10 D) 6 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
