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 88 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) 1 B) 2 C) 5 D) 3 2 / 25 2. Agar f(g(x))=x²-6x+6 va f(x)=x−3 funksiyalar berilgan bo`lsa, g(x) funksiya ko`rinishini aniqlang. A) (x-3)² B) x²+12x+3 C) (x+3)² D) x²-12-3 3 / 25 3. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 0 B) 1 C) -2 D) -1 4 / 25 4. Agar ni toping. A) 0.8 B) 0.9 C) 0.96 D) 0.81 5 / 25 5. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 0 B) -1 C) 1 D) 3 6 / 25 6. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) √2 B) 1 C) 2√2 D) √3 7 / 25 7. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -3 B) -2 C) 2 D) 3 8 / 25 8. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -25 B) -24 C) -26 D) -27 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) 6 D) 8 10 / 25 10. ABCD to‘rtburchakda АВ = CD = 9 va bu to‘rtburchakka radiusi 4 ga teng aylana ichki chizilgan. ABCD to‘rtburchak yuzini toping. A) 72 B) 144 C) 81 D) 36 11 / 25 11. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x ≤ 3 B) x<0 C) x>3 D) 3 ≤ x 12 / 25 12. 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 13 / 25 13. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 8? + 2 B) 10? + 2 C) 6? + 2 D) 4? + 2 14 / 25 14. Agar bo‘lsa, |xyz| ni toping. A) 500 B) 800 C) 100 D) 1000 15 / 25 15. bo‘lsa ? ning qiymatini toping. A) 40 B) 25 C) 37 D) 12 16 / 25 16. Agar A) 0 B) 3π/2 C) π/2 D) π 17 / 25 17. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 8 B) 0 C) 16 D) 12 18 / 25 18. 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) 2 C) 3 D) 9 19 / 25 19. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 3 B) 5 C) 2 D) 8 20 / 25 20. 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) 11 B) 10 C) 8 D) 9 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? − 30 B) −6? + 42 C) −6? + 6 D) 6? − 24 22 / 25 22. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 20x+48 B) x+25 C) 20x+50 D) 0 23 / 25 23. 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 24 / 25 24. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) √21 B) 4√21 C) 35 D) 2√21 25 / 25 25. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) sin18° B) 2cos² 9° C) 1 D) tg9° 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
