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 149 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Savollar 12345678910111213141516171819202122232425 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster 10-sinf Matematika olimpiada №3 2021-2022 o'quv yili viloyat bosqichida tushgan savollar. 1 / 25 1. tengsizlik qiymatlarida o‘rinli bo‘ladigan ning barcha qiymatlarini toping. A) (8;19) B) [2;8] C) (2;8) D) (-2;8) 2 / 25 2. ifodani soddalshtiring A) 2sinα B) 2cosα C) cosα D) 1 3 / 25 3. 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) 9 D) 8 4 / 25 4. 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 5 / 25 5. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 5 B) 7 C) 6 D) 4 6 / 25 6. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 3 B) -1 C) 0 D) 1 7 / 25 7. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 8? + 2 B) 6? + 2 C) 4? + 2 D) 10? + 2 8 / 25 8. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -26 C) -25 D) -24 9 / 25 9. x, y sonlari (x² + 1)(y² + 1) + 2(x − y)(1 − xy) = 4(1 + xy) tenglikni qanoatlantiradi |1 + x| ∙ |1 − y| ni toping. A) 2 B) 1 C) 9 D) 3 10 / 25 10. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) √21 B) 35 C) 4√21 D) 2√21 11 / 25 11. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 8 B) 12 C) 16 D) 0 12 / 25 12. 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? − 30 C) −6? + 6 D) −6? + 42 13 / 25 13. bo‘lsa ? ning qiymatini toping. A) 12 B) 40 C) 25 D) 37 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) 1 B) √3 C) 2√2 D) √2 15 / 25 15. Tenglama nechta yechimga ega? A) 5 B) 2 C) 4 D) 3 16 / 25 16. Agar ni toping. A) 0.9 B) 0.8 C) 0.96 D) 0.81 17 / 25 17. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 1 B) -1 C) 0 D) -2 18 / 25 18. Agar bo‘lsa, |xyz| ni toping. A) 500 B) 800 C) 1000 D) 100 19 / 25 19. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) 3 B) 2 C) -2 D) -3 20 / 25 20. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 1 B) sin18° C) 2cos² 9° D) tg9° 21 / 25 21. 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) 144 C) 36 D) 72 22 / 25 22. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x<0 B) x ≤ 3 C) x>3 D) 3 ≤ x 23 / 25 23. 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) 6 C) 5 D) 8 24 / 25 24. Agar A) π B) π/2 C) 0 D) 3π/2 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) 150° B) 90° C) 135° D) 120° 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
