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. Agar ikki xonali sonning o‘ng tomoniga 6 raqami yozib qo‘yilsa, berilgan sonning yarmi bilan 500 ning yig‘indisiga teng son hosil bo‘ladi. Bu qanday ikki xonali son? A) 30 B) 52 C) 43 D) 60 2 / 25 2. 7(kx−2)=7k+2(x−8) tenglama k ning qanday qiymatlarida ildizga ega bo`lmaydi? A) (−∞; 2/7) B) k=2/7 C) ∅ D) (2/7; ∞) 3 / 25 3. 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 4 / 25 4. 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) 144 C) 72 D) 81 5 / 25 5. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -2 B) 3 C) -3 D) 2 6 / 25 6. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) -2 B) 1 C) 0 D) -1 7 / 25 7. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 10 B) 9 C) 15 D) 12 8 / 25 8. Agar ni toping. A) 0.8 B) 0.81 C) 0.9 D) 0.96 9 / 25 9. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 16 B) 15 C) 14 D) 17 10 / 25 10. englikni qanoatlantiradigan ? ning eng kichik qiymatini toping. A) 3 B) -1 C) 1 D) 0 11 / 25 11. 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) 11 D) 10 12 / 25 12. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -27 B) -25 C) -26 D) -24 13 / 25 13. Tenglama nechta yechimga ega? A) 3 B) 4 C) 5 D) 2 14 / 25 14. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 5 B) 2 C) 8 D) 3 15 / 25 15. Tenglamalar sistemasi nechta yechimga ega? A) 8 B) 12 C) 6 D) 10 16 / 25 16. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) sin18° B) tg9° C) 2cos² 9° D) 1 17 / 25 17. 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) 5√3/12 C) 5√2/5 D) 2/5 18 / 25 18. 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 19 / 25 19. a,b,c−ABC uchburchakning tomonlari. Agar a4 + b4 + c4 + 32 = 2(a²b² + b²c² + a²c²) bo‘lsa, ABC uchburchak yuzini toping. A) 2√2 B) √3 C) √2 D) 1 20 / 25 20. 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 21 / 25 21. ?² + |? − 3| ≤ |?2 + ?| − 3 tengsizlikni yeching. A) x>3 B) x<0 C) 3 ≤ x D) x ≤ 3 22 / 25 22. ?(?) = ?5 − 7?4 + 3?3 − ? + 2 ko‘phadni ?2 + ? ga bo‘lgandagi qoldiqni aniqlang. A) 10? + 2 B) 6? + 2 C) 4? + 2 D) 8? + 2 23 / 25 23. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 6 B) 7 C) 4 D) 5 24 / 25 24. Agar A) π/2 B) 3π/2 C) 0 D) π 25 / 25 25. Agar bo‘lsa, |xyz| ni toping. A) 1000 B) 100 C) 800 D) 500 0% Testni qayta ishga tushiring Author: InfoMaster Matematika olimpiada
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