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 49 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 y ning qanday qiymatlarida o‘rinli. A) 3 B) 9 C) 729 D) 343 2 / 25 2. 7(kx−2)=7k+2(x−8) tenglama k ning qanday qiymatlarida ildizga ega bo`lmaydi? A) (−∞; 2/7) B) (2/7; ∞) C) ∅ D) k=2/7 3 / 25 3. 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) 8 B) 15 C) 10 D) 9 4 / 25 4. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 5 B) 2 C) 3 D) 8 5 / 25 5. Agar bo‘lsa, |xyz| ni toping. A) 800 B) 100 C) 1000 D) 500 6 / 25 6. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 2cos² 9° B) sin18° C) tg9° D) 1 7 / 25 7. 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) 135° B) 90° C) 120° D) 150° 8 / 25 8. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 10? + 2 B) 8? + 2 C) 4? + 2 D) 6? + 2 9 / 25 9. 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) 10 B) 9 C) 11 D) 8 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? + 42 B) 6? − 24 C) 6? − 30 D) −6? + 6 11 / 25 11. Tenglamalar sistemasi nechta yechimga ega? A) 6 B) 8 C) 12 D) 10 12 / 25 12. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 1 B) 0 C) -1 D) -2 13 / 25 13. Tenglama nechta yechimga ega? A) 4 B) 2 C) 3 D) 5 14 / 25 14. Agar A) 3π/2 B) 0 C) π/2 D) π 15 / 25 15. Agar ni toping. A) 0.8 B) 0.81 C) 0.96 D) 0.9 16 / 25 16. 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 17 / 25 17. 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) 3 C) 9 D) 2 18 / 25 18. 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 19 / 25 19. bo‘lsa ? ning qiymatini toping. A) 40 B) 12 C) 25 D) 37 20 / 25 20. 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 21 / 25 21. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 0 B) 20x+50 C) 20x+48 D) x+25 22 / 25 22. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 9 B) 12 C) 10 D) 15 23 / 25 23. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -3 B) 3 C) 2 D) -2 24 / 25 24. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -26 B) -25 C) -27 D) -24 25 / 25 25. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x<0 B) 3 ≤ x C) x ≤ 3 D) x>3 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
