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 121 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. a ning qanday qiymatlarida kvadrat tenglamaning ildizlari yig‘indisi manfiy bo‘ladi? A) a<1 B) a<0 C) 0<a<2 D) 0<a<1 2 / 25 2. 2x²-2ax+1=0 kvadrat tenglama ildizlari uchun tengliko`rinlibo`lsa, a ni toping. A) 7 B) -1 C) 1 D) -3 3 / 25 3. Agar ni toping. A) 0.8 B) 0.81 C) 0.96 D) 0.9 4 / 25 4. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 7 B) 4 C) 5 D) 6 5 / 25 5. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 35 B) 2√21 C) 4√21 D) √21 6 / 25 6. Agar A) 3π/2 B) π C) 0 D) π/2 7 / 25 7. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) √3 B) 1 C) √2 D) 2√2 8 / 25 8. 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) 72 C) 81 D) 144 9 / 25 9. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -24 C) -26 D) -25 10 / 25 10. 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) 9 C) 10 D) 8 11 / 25 11. 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) 15 D) 9 12 / 25 12. Tenglama nechta yechimga ega? A) 4 B) 3 C) 2 D) 5 13 / 25 13. 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) 2/5 C) 5√2/5 D) 5√3/12 14 / 25 14. 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) 11 C) 9 D) 8 15 / 25 15. 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 16 / 25 16. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 16 B) 15 C) 14 D) 17 17 / 25 17. 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) 120° C) 150° D) 135° 18 / 25 18. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 20x+48 B) x+25 C) 0 D) 20x+50 19 / 25 19. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 5 B) 3 C) 8 D) 2 20 / 25 20. bo‘lsa ? ning qiymatini toping. A) 12 B) 37 C) 25 D) 40 21 / 25 21. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 1 B) -1 C) -2 D) 0 22 / 25 22. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 4? + 2 B) 8? + 2 C) 6? + 2 D) 10? + 2 23 / 25 23. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x>3 B) 3 ≤ x C) x ≤ 3 D) x<0 24 / 25 24. Tenglamalar sistemasi nechta yechimga ega? A) 8 B) 10 C) 12 D) 6 25 / 25 25. 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) 8 C) 6 D) 5 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz