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 113 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. ABCD trapetsiyaning balandligi 10 ga teng va asoslari AD:BC=7:3 shartni qanoattlantiradi. Trapetsiyaning diagonallari kesishgan nuqtadan katta asosgacha bo‘lgan masofani toping. A) 7 B) 9 C) 8 D) 10 2 / 25 2. [2x-1]=x tenglamani butun yechimlar sonini toping. Bu yerda [a]− a sonining butun qismi. A) 2 ta B) 1 ta C) 3 ta D) 4 ta 3 / 25 3. Tenglamalar sistemasi nechta yechimga ega? A) 8 B) 10 C) 6 D) 12 4 / 25 4. Tenglama nechta yechimga ega? A) 2 B) 4 C) 3 D) 5 5 / 25 5. 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) 7 C) 6 D) 8 6 / 25 6. ?(? − 1) = 2?(5? + 4) va ?(2? − 1) = 4? + 4 bo‘lsa, ?(?) ni toping A) 0 B) x+25 C) 20x+48 D) 20x+50 7 / 25 7. bo‘lsa ? ning qiymatini toping. A) 12 B) 25 C) 40 D) 37 8 / 25 8. 102022 − 22021 ayirmani 24 ga bo‘lgandagi qoldiqni toping. A) 0 B) 12 C) 16 D) 8 9 / 25 9. Tenglamani yeching: 3x+3 + 8·3x+2 = 33 A) 0 B) -2 C) -1 D) 1 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? + 6 B) −6? + 42 C) 6? − 24 D) 6? − 30 11 / 25 11. ABC uchburchakning AC tomonida D nuqta olingan, bunda ∠ABC = ∠BDC. Agar AD = 10, CD = 8 bo‘lsa, BC ni toping. A) 9 B) 10 C) 15 D) 12 12 / 25 12. ?/? kasr (?, ? −natural sonlar)— qisqarmas kasr va (7?+6?)/(3?+2?) kasr esa qisqaradi. Ushbu kasr qanday songa qisqaradi? A) 8 B) 2 C) 5 D) 3 13 / 25 13. (?² − 2? + 3)(?² + 6? + 12) = 6 bo‘lsa, ? + ? ni toping. A) -2 B) 2 C) -3 D) 3 14 / 25 14. 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) 8 B) 10 C) 9 D) 15 15 / 25 15. Agar ni toping. A) 0.96 B) 0.9 C) 0.8 D) 0.81 16 / 25 16. 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) 8 C) 9 D) 11 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) 120° B) 135° C) 90° D) 150° 18 / 25 18. Agar bo‘lsa, |xyz| ni toping. A) 1000 B) 100 C) 800 D) 500 19 / 25 19. Agar A) π/2 B) 3π/2 C) 0 D) π 20 / 25 20. tenglama nechta yechimga ega, agar x ∈ (0; 50)? A) 14 B) 15 C) 16 D) 17 21 / 25 21. Agar bo‘lsa, ?(0) − ?(5) ayirmani toping. A) -24 B) -26 C) -25 D) -27 22 / 25 22. 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) 81 D) 72 23 / 25 23. x, y sonlari (x² + 1)(y² + 1) + 2(x − y)(1 − xy) = 4(1 + xy) tenglikni qanoatlantiradi |1 + x| ∙ |1 − y| ni toping. A) 2 B) 9 C) 1 D) 3 24 / 25 24. Agar ? > 0, ? + ?² = 7,25; ?² − ? = 2 va ?² = √(? − 1) ∙ √(2 − ?) bo‘lsa, ?(√(? − 1) + √(2 − ?)) ning qiymatini toping. A) 4 B) 6 C) 7 D) 5 25 / 25 25. Soddalashtiring: (4cos² 9° −3)(4cos² 27° −3)·ctg9° A) 2cos² 9° B) tg9° C) 1 D) sin18° 0% Testni qayta ishga tushiring Author: InfoMaster Foydali bo'lsa mamnunmiz
