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 107 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. Hisoblang: A) 4 B) 1 C) 3 D) 2 2 / 25 2. Beshta a1, a2, a3, a4, a5 tub sonlar ayirmasi 6 ga teng bo`lgan arifmetik progressiyani tashkil qiladi. 2a2+a3 ni toping. A) 31 B) 43 C) 39 D) 40 3 / 25 3. 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 4 / 25 4. Agar bo‘lsa, |xyz| ni toping. A) 100 B) 500 C) 1000 D) 800 5 / 25 5. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 10? + 2 B) 6? + 2 C) 8? + 2 D) 4? + 2 6 / 25 6. 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) 144 C) 72 D) 81 7 / 25 7. 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 8 / 25 8. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -25 B) -24 C) -26 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) 10 B) 9 C) 15 D) 12 10 / 25 10. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 2 B) 8 C) 5 D) 3 11 / 25 11. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x<0 B) 3 ≤ x C) x>3 D) x ≤ 3 12 / 25 12. 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√6/24 B) 5√2/5 C) 5√3/12 D) 2/5 13 / 25 13. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 15 B) 17 C) 16 D) 14 14 / 25 14. 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) √2 C) 1 D) √3 15 / 25 15. 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) 8 B) 5 C) 6 D) 7 16 / 25 16. 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° 17 / 25 17. 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 18 / 25 18. Agar ni toping. A) 0.9 B) 0.81 C) 0.96 D) 0.8 19 / 25 19. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 16 B) 0 C) 12 D) 8 20 / 25 20. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) tg9° B) 2cos² 9° C) sin18° D) 1 21 / 25 21. bo‘lsa ? ning qiymatini toping. A) 40 B) 12 C) 25 D) 37 22 / 25 22. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) 2 B) 3 C) -2 D) -3 23 / 25 23. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 20x+48 B) x+25 C) 0 D) 20x+50 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) 9 B) 11 C) 10 D) 8 25 / 25 25. 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) 15 B) 8 C) 10 D) 9 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
