10-sinf Matematika olimpiada №3 Oktabr 3, 2022Oktabr 3, 2022 da chop etilgan InfoMaster tomonidan 10-sinf Matematika olimpiada №3 ga 7 fikr bildirilgan 0 12345678910111213141516171819202122232425 Vaqtingiz tugadi! 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) (2;8) B) (8;19) C) [2;8] D) (-2;8) 2 / 25 2. Ta`lim muassasida barcha o`quvchilar kamida bitta ingliz yoki nemis tilida so`zlashа oladilar. Ayrimlari ikkala tilni ham biladilar. O`quvchilarning 85%i ingliz tilini, 75%i nemis tilini biladilar. Ikkala tilni ham biladigan o`quvchilar barcha o`quvchilarning necha % ini tashkil etadi. A) 60% B) 70% C) 50% D) 65% 3 / 25 3. 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) 8 D) 6 4 / 25 4. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -25 C) -26 D) -24 5 / 25 5. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x<0 B) 3 ≤ x C) x>3 D) x ≤ 3 6 / 25 6. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 1 B) 0 C) -1 D) -2 7 / 25 7. 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) 8 D) 10 8 / 25 8. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 4? + 2 B) 6? + 2 C) 10? + 2 D) 8? + 2 9 / 25 9. 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) 120° C) 90° D) 150° 10 / 25 10. Uchburchak tomonlarining uzunliklari berilgan tenglamaning ildizlariga mos keladi. ?³ − 24?² + 183? − 440 = 0. Uchburchakning yuzini hisoblang. A) 4√21 B) 35 C) 2√21 D) √21 11 / 25 11. Tenglamalar sistemasi nechta yechimga ega? A) 10 B) 8 C) 6 D) 12 12 / 25 12. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) √2 B) √3 C) 1 D) 2√2 13 / 25 13. 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? + 6 C) 6? − 30 D) −6? + 42 14 / 25 14. ABCD to‘rtburchakda АВ = CD = 9 va bu to‘rtburchakka radiusi 4 ga teng aylana ichki chizilgan. ABCD to‘rtburchak yuzini toping. A) 72 B) 81 C) 144 D) 36 15 / 25 15. Agar ni toping. A) 0.96 B) 0.8 C) 0.81 D) 0.9 16 / 25 16. Tenglama nechta yechimga ega? A) 4 B) 5 C) 3 D) 2 17 / 25 17. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) x+25 B) 20x+48 C) 20x+50 D) 0 18 / 25 18. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 1 B) tg9° C) 2cos² 9° D) sin18° 19 / 25 19. Agar A) π/2 B) 3π/2 C) π D) 0 20 / 25 20. Agar bo‘lsa, |xyz| ni toping. A) 500 B) 1000 C) 800 D) 100 21 / 25 21. 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 22 / 25 22. 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) 1 C) 2 D) 3 23 / 25 23. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 5 B) 8 C) 3 D) 2 24 / 25 24. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 17 B) 15 C) 16 D) 14 25 / 25 25. 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) 2/5 C) 5√6/24 D) 5√2/5 0% Testni qayta ishga tushiring Author: InfoMaster Matematika olimpiada
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