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 1 Ko'rish 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. tenglamalar sistemasining yechimlari bo‘lsin. ayirmani toping. A) 4 B) -3 C) -4 D) -5 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) 39 C) 40 D) 43 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) 9 B) 8 C) 10 D) 11 4 / 25 4. 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) 10 C) 8 D) 9 5 / 25 5. Tenglama nechta yechimga ega? A) 5 B) 3 C) 4 D) 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) 81 C) 72 D) 144 7 / 25 7. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x>3 B) x ≤ 3 C) x<0 D) 3 ≤ x 8 / 25 8. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 2√21 B) 35 C) 4√21 D) √21 9 / 25 9. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) 3 B) 2 C) -2 D) -3 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. Agar ni toping. A) 0.96 B) 0.9 C) 0.81 D) 0.8 12 / 25 12. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 14 B) 16 C) 15 D) 17 13 / 25 13. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 16 B) 0 C) 8 D) 12 14 / 25 14. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -26 B) -27 C) -25 D) -24 15 / 25 15. 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 16 / 25 16. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 15 B) 9 C) 10 D) 12 17 / 25 17. Agar bo‘lsa, |xyz| ni toping. A) 1000 B) 500 C) 100 D) 800 18 / 25 18. 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) 2/5 B) 5√2/5 C) 5√3/12 D) 5√6/24 19 / 25 19. Agar A) 0 B) π C) 3π/2 D) π/2 20 / 25 20. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) -1 B) 0 C) 3 D) 1 21 / 25 21. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) -1 B) 1 C) 0 D) -2 22 / 25 22. bo‘lsa ? ning qiymatini toping. A) 40 B) 12 C) 25 D) 37 23 / 25 23. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 7 B) 4 C) 6 D) 5 24 / 25 24. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 6? + 2 B) 10? + 2 C) 4? + 2 D) 8? + 2 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) 5 B) 8 C) 7 D) 6 0% Testni qayta ishga tushiring Tomonidan Wordpress Quiz plugin Author: InfoMaster Foydali bo'lsa mamnunmiz
