10-sinf Matematika olimpiada №3 Oktabr 3, 2022Oktabr 3, 2022 da chop etilgan InfoMaster tomonidan 10-sinf Matematika olimpiada №3 ga fikr bildirilmagan. 0 12345678910111213141516171819202122232425 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster 10-sinf Matematika olimpiada №3 2021-2022 o'quv yili viloyat bosqichida tushgan savollar. 1 / 25 1. Agar bo‘lsa, ni toping. A) -5 B) -1 C) -7 D) -3 2 / 25 2. ifodani soddalshtiring A) 1 B) cosα C) 2cosα D) 2sinα 3 / 25 3. Tenglamalar sistemasi nechta yechimga ega? A) 8 B) 6 C) 12 D) 10 4 / 25 4. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 12 B) 10 C) 9 D) 15 5 / 25 5. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 5 B) 2 C) 8 D) 3 6 / 25 6. Agar ni toping. A) 0.96 B) 0.8 C) 0.81 D) 0.9 7 / 25 7. Agar A) 3π/2 B) π/2 C) 0 D) π 8 / 25 8. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x>3 B) 3 ≤ x C) x ≤ 3 D) x<0 9 / 25 9. bo‘lsa ? ning qiymatini toping. A) 40 B) 12 C) 25 D) 37 10 / 25 10. 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) 2 C) 9 D) 1 11 / 25 11. 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 12 / 25 12. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 4 B) 6 C) 5 D) 7 13 / 25 13. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 0 B) 8 C) 16 D) 12 14 / 25 14. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 0 B) 3 C) -1 D) 1 15 / 25 15. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) sin18° B) tg9° C) 1 D) 2cos² 9° 16 / 25 16. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -24 B) -26 C) -27 D) -25 17 / 25 17. 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) 15 C) 8 D) 9 18 / 25 18. Agar bo‘lsa, |xyz| ni toping. A) 800 B) 1000 C) 500 D) 100 19 / 25 19. Agar ?1 = 6? − 6 va у2//?1 hamda у2 to‘g‘ri chiziq ?(6; 6) nuqtadan o‘tsa, ?2 ni toping. A) 6? − 24 B) −6? + 42 C) −6? + 6 D) 6? − 30 20 / 25 20. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 10? + 2 B) 6? + 2 C) 4? + 2 D) 8? + 2 21 / 25 21. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 20x+48 B) 20x+50 C) 0 D) x+25 22 / 25 22. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 4√21 B) 35 C) 2√21 D) √21 23 / 25 23. 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) 8 D) 11 24 / 25 24. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) 3 B) 2 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) 120° B) 135° C) 90° D) 150° 0% Testni qayta ishga tushiring Author: InfoMaster Matematika olimpiada