10-sinf Matematika olimpiada №3 Oktabr 3, 2022Avgust 19, 2024 da chop etilgan InfoMaster tomonidan 10-sinf Matematika olimpiada №3 ga 1 fikr bildirilgan 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 ikkita teng tomonli uchburchakning tomonlari uchun munosabat o‘rinli bo‘lsa, katta uchburchak yuzini kichik uchburchak yuziga nisbatini toping. A) 8 B) 7 C) 6 D) 9 2 / 25 2. Korxona xom−ashyuo qimmatlashgani sababli maxsulot narxi 20% ga oshirildi, biroq mijojlar soni 20 % ga kamayib ketdi. Shunda korxona narxni qanchadir foizga kamaytirib mijozlari sonini 25% ga oshirishga erishdi. Shundan keyin korxona tushumi narx o`zgarmasdan oldingi tushumga qaraganda 8% ga oshgan bo`lsa, yangi narx necha foizga kamaytirilgan? A) 10 B) 4 C) 20 D) 12 3 / 25 3. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -25 B) -24 C) -26 D) -27 4 / 25 4. Tenglamalar sistemasi nechta yechimga ega? A) 6 B) 12 C) 8 D) 10 5 / 25 5. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 20x+48 B) 20x+50 C) 0 D) x+25 6 / 25 6. Agar ni toping. A) 0.9 B) 0.8 C) 0.81 D) 0.96 7 / 25 7. 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? + 6 C) 6? − 30 D) 6? − 24 8 / 25 8. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 8 B) 0 C) 16 D) 12 9 / 25 9. 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) 5 C) 6 D) 8 10 / 25 10. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) √2 B) 1 C) √3 D) 2√2 11 / 25 11. bo‘lsa ? ning qiymatini toping. A) 12 B) 40 C) 37 D) 25 12 / 25 12. 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) 11 C) 10 D) 8 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√2/5 B) 2/5 C) 5√3/12 D) 5√6/24 14 / 25 14. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 14 B) 17 C) 15 D) 16 15 / 25 15. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) sin18° B) tg9° C) 2cos² 9° D) 1 16 / 25 16. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 6? + 2 B) 8? + 2 C) 4? + 2 D) 10? + 2 17 / 25 17. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 2√21 B) 35 C) √21 D) 4√21 18 / 25 18. x, y sonlari (x² + 1)(y² + 1) + 2(x − y)(1 − xy) = 4(1 + xy) tenglikni qanoatlantiradi |1 + x| ∙ |1 − y| ni toping. A) 9 B) 2 C) 1 D) 3 19 / 25 19. 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 20 / 25 20. Tenglama nechta yechimga ega? A) 3 B) 2 C) 4 D) 5 21 / 25 21. 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) 150° C) 120° D) 90° 22 / 25 22. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 4 B) 5 C) 7 D) 6 23 / 25 23. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -2 B) 3 C) -3 D) 2 24 / 25 24. Agar A) π B) 0 C) π/2 D) 3π/2 25 / 25 25. Agar bo‘lsa, |xyz| ni toping. A) 800 B) 500 C) 1000 D) 100 0% Testni qayta ishga tushiring tomonidan Wordpress Quiz plugin Matematika olimpiada, Viloyat 2021-2022 o'quv yili
