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+ 1 10 · 17 + 1 17 · 24 + 1 24 · 31 + 1 31 · 38 A) 5 114 B) 7 104 C) 5 104 D) 7 114 16. (98-12-62)* Hisoblang. 1 1 · 2 + 1 2 · 3 + 1 3 · 4 + · · · + 1 999 · 1000 13 A) 0, 750 B) 1, 125 C) 0, 998 D) 0, 999 Yechish: Yuqorida keltirilgan yechish usulidan foydalanamiz. Etibor qilgan bo’lsangiz yig’indida birinchi qo’shiluvchi minus oxirgisi qoladi. Ixtiy- oriy n ∈ N uchun 1 n(n + 1) = 1 n − 1 n + 1 dan (1− 1 2 )+( 1 2 − 1 3 )+( 1 3 − 1 4 )+· · ·+( 1 999 − 1 1000 ) = = 1 − 1 1000 = 999 1000 = 0, 999 ni olamiz. Javob: 0, 999 (D). 17. (00-3-15)* Hisoblang. 1 1 · 3 + 1 3 · 5 + 1 5 · 7 + · · · + 1 13 · 15 A) 11 15 B) 7 30 C) 8 15 D) 7 15 18. (00-8-57)* Hisoblang. 1 1 · 2 + 1 2 · 3 + 1 3 · 4 + · · · + 1 99 · 100 A) 1 9 B) 1 10 C) 1 100 D) 99 100 19. (03-7-43)* Hisoblang. 2 5 · 7 + 2 7 · 9 + 2 9 · 11 + · · · + 2 73 · 75 A) 16 75 B) 28 75 C) 1 5 D) 14 75 20. (03-8-29)* Hisoblang. 1 + 1 10 · 11 + 1 11 · 12 + 1 12 · 13 + 1 13 · 14 + 1 14 · 15 A) 1 3 80 B) 1, 16 C) 1 1 30 D) 1 7 80 21. (03-1-9)* Hisoblang. 1 8 + 1 24 + 1 48 + 1 80 A) 0, 1 B) 0, 2 C) 0, 4 D) 0, 6 Yechish: Yuqorida yechilgan 11-misol usulidan foydalanamiz: 1 8 + 1 24 + 1 48 + 1 80 = 1 2 · 4 + 1 4 · 6 + 1 6 · 8 + 1 8 · 10 = 1 2 ( 1 2 − 1 4 + 1 4 − 1 6 + 1 6 − 1 8 + 1 8 − − 1 10 ) = 1 2 ( 1 2 − 1 10 ) = 1 2 · 4 10 = 1 5 = 0, 2. Javob: 0, 2 (B). 22. (00-2-4)* Hisoblang. 1 15 + 1 35 + 1 63 + 1 99 + 1 143 + 1 195 A) 4 15 B) 7 15 C) 17 45 D) 2 15 23. (99-5-1)* Hisoblang. 1 12 + 1 20 + 1 30 + 1 42 + · · · + 1 182 A) 11 42 B) 10 33 C) 1 4 D) 12 35 24. (00-9-10)* Hisoblang. 1 15 + 1 35 + 1 63 + 1 99 + · · · + 1 255 A) 7 51 B) 2 15 C) 2 25 D) 3 35 25. (96-11-11) Hisoblang. ³ − 1 3 ´ · 2 7 : 5 42 A) 4 5 B) 5 441 C) − 4 5 D) 10 882 Yechish: Ko’paytirish (4-ga qarang) va bo’lish (7-ga qarang) qoidalaridan foydalanamiz: ³ − 1 3 ´ · 2 7 : 5 42 = − 1 3 · 2 7 · 42 5 = − 1 3 · 2 7 · 2 · 3 · 7 5 = − 4 5 . Javob: − 4 5 (C). 26. (96-12-10) Hisoblang. − 1 3 · ³ − 2 7 ´ : 5 42 A) 5 441 B) 4 5 C) − 5 441 D) − 4 5 27. Hisoblang. 1 17 : ³ − 5 51 ´ · 4 15 A) 5 441 B) 4 25 C) − 5 24 D) − 4 25 28. Hisoblang. 3 8 : 3 4 ³ − 2 5 ´ : 1 20 A) 5 4 B) 4 C) −4 D) − 4 25 29. (98-7-5) Hisoblang. 243 : (9 : 11) A) 27 B) 2 5 11 C) 11 27 D) 297 30. Hisoblang. 3 8 : 3 4 − 1 3 · ³ − 2 7 ´ : ³ − 5 42 ´ A) 7 10 B) − 3 10 C) 5 42 D) 3 35 Yechish: Ko’paytirish (4-ga qarang) va bo’lish (7-ga qarang) qoidalaridan foydalanamiz: 3 8 : 3 4 − 1 3 · ³ − 2 7 ´ : ³ − 5 42 ´ = 3 8 · 4 3 − 1 3 · 2 7 · 42 5 = = 1 2 − 4 5 = 5 − 8 10 = − 3 10 . Javob: − 3 10 (B). 14 31. Hisoblang. 3 8 : 3 4 − 1 3 : ³ − 2 9 ´ A) 10 B) −1 C) 2 D) 3, 5 32. Hisoblang. 3 8 : 3 4 + 1 3 · ³ − 2 7 ´ : ³ − 5 42 ´ A) 1 1 10 B) − 3 10 C) 5 42 D) 3 35 33. Hisoblang. 7 8 : 7 4 + 1 5 · ³ − 2 7 ´ : ³ − 4 35 ´ A) 10 B) 0 C) 3 35 D) 3 34. Hisoblang. 5 8 : 5 4 + 1 3 · ³ − 2 7 ´ : ³ − 35 42 ´ A) 1 5 6 B) 9 10 C) 5 6 D) 7 42 35. (96-9-58) 3 4 va 8 9 sonlari orasida maxraji 36 ga teng bo’lgan nechta kasr son bor? A) 1 B) 2 C) 3 D) 4 Yechish: Berilgan kasrlarni 36 maxrajga kelti- ramiz. 3 4 = 27 36 va 8 9 = 32 36 . Bu sonlar orasida maxraji 36 ga teng bo’lgan 28 36 , 29 36 , 30 36 , 31 36 son- lar bor. Javob: 4 (D). 36. 3 5 va 11 15 sonlari orasida maxraji 45 ga teng bo’lgan nechta kasr son bor? A) 5 B) 2 C) 3 D) 4 37. 3 4 va 9 10 sonlari orasida maxraji 40 ga teng bo’lgan nechta kasr son bor? A) 6 B) 5 C) 3 D) 4 38. (96-1-8) 9 11 va 1 sonlari orasida maxraji 33 ga teng bo’lgan nechta kasr son bor? A) 2 B) 1 C) 5 D) 6 39. (96-10-8) 2 3 dan katta va 5 6 dan kichik bo’lgan, maxraji 30 ga teng bo’lgan nechta kasr mavjud? A) 1 B) 2 C) 4 D) 5 1.2.2 Aralash kasrlar Bizga a n , a ≥ n noto’g’ri kasr berilgan bo’lsin. Qoldiqli bo’lish (1.1.5-dagi 1-ga qarang) qoidasiga ko’ra, shun- day m natural va r (0 ≤ r < n) butun sonlar mavjud bo’lib, bular uchun a = nm + r tenglik o’rinli bo’ladi. U holda a n = nm + r n = nm n + r n = m + r n . Demak, a n noto’g’ri kasrni natural son m bilan to’g’ri kasr r n ning yig’indisi shaklida yozish mumkin. Bunday amal noto’g’ri kasrdan butun ajratish deyiladi. Masalan, 12 5 = 10 + 2 5 = 10 5 + 2 5 = 2 + 2 5 . Natural son bilan to’g’ri kasrning yig’indisini qo’shish amalisiz yozish qabul qilingan, ya’ni 2 + 2 5 o’rniga 2 2 5 ko’rinishda yozi- ladi. Bu son aralash kasr deyiladi. Endi musbat kasr- larni taqqoslash haqida qoidalar beramiz. 1. Bir xil maxrajli a b va c b kasrlarning surati kattasi katta bo’ladi, ya’ni a > c ⇐⇒ a b > c b . 2. Suratlari bir xil a b va a c kasrlarning maxraji kattasi kichik bo’ladi, ya’ni b < c ⇐⇒ a b > a c . 3. Ikki aralash kasrning butun qismi kattasi katta bo’ladi. Agar ularning butun qism- lari teng bo’lsa, u holda kasr qismi kattasi katta bo’ladi. 4. 0 < a < b < c ⇐⇒ 1 a > 1 b > 1 c . 5. a soniga teskari son 1 a . Masalan, a = 0, 8 bo’lsa, unga teskari son 1 0, 8 = 1, 25. 6. a soniga qarama-qarshi son − a. Masalan, a = 0, 8 bo’lsa, unga qarama-qarshi son − 0, 8. 1. (97-5-9) Amalni bajaring. 1 3 5 − 3 1 5 A) −1 2 5 B) 1 2 5 C) 1 3 5 D) −1 3 5 Yechish: Aralash kasrlarni ayirishda butun qis- midan butun qismi, kasr qismidan kasr qismi ayri- ladi. Demak, 1 3 5 − 3 1 5 = (1 − 3) + ( 3 5 − 1 5 ) = = −2 + 2 5 = −(2 − 2 5 ) = − 10 − 2 5 = − 8 5 = −1 3 5 . Javob: −1 3 5 . (D). 2. (97-9-9) Amalni bajaring. 3 4 7 − 5 2 7 A) −1 5 7 B) 1 4 7 C) 1 5 7 D) − 4 7 3. Amalni bajaring. 3 4 9 − 2 5 18 A) −1 1 6 B) 1 5 6 C) 1 1 6 D) −1 1 18 4. Amalni bajaring. 2011 1 9 − 2010 11 18 A) −0, 1 B) 0, 5 C) 0, 6 D) −0, 5 5. (96-1-7) 0, 6 ga teskari sonni toping. A) −0, 6 B) 1 2 3 C) 0, 4 D) −6 Yechish: Teskari son ta’rifiga ko’ra (5-ga qarang), 0, 6 = 6 10 ga teskari son 10 6 = 5 3 = 1 2 3 dir. Javob: 1 2 3 (B). 15 6. (96-9-57) 0, 8 songa teskari sonni toping. A) −0, 8 B) 8 C) − 5 4 D) 1, 25 7. (96-10-7) −1, 5 soniga teskari sonni toping. A) 1, 5 B) −0, 75 C) − 2 3 D) 2 3 8. (00-2-6) 11 25 va 4 6 11 sonlariga teskari sonlar ko’- paytmasini toping. A) 1 2 B) 1 C) 3 4 D) 2 9. (06-111-15) O’zaro teskari sonlarni aniqlang: 1) 3 − √ 2 va 3 + √ 2; 2) √ 5 3 va 3 √ 5 5 ; 3) 2 √ 3 5 va 5 √ 3 6 ; 4) √ 2 + 1 va √ 2 − 1; A) 1; 2; 3 B) 1; 3; 4 C) 1; 3 D) 2; 3; 4 10. 2, 5 ga qarama-qarshi sonni toping. A) 0, 4 B) −0, 4 C) −2, 5 D) 2 Yechish: Qarama-qarshi son ta’rifiga ko’ra (6- ga qarang), 2, 5 ga qarama-qarshi son −2, 5 dir. Javob: −2, 5 (C). 11. −1, 6 ga qarama-qarshi sonni toping. A) 0, 125 B) 1, 6 C) 0, 8 D) 16 12. 0, 4 ga qarama-qarshi bo’lgan songa teskari sonni toping. A) 0, 4 B) −0, 4 C) −2, 5 D) 2 13. (03-1-56) 0, 8 ga teskari bo’lgan songa qarama- qarshi sonni toping. A) −0, 8 B) 1, 25 C) −1, 25 D) −1, 2 14. (98-3-5) Agar a = 5 11 , b = 3 7 , c = 6 13 bo’lsa, a, b va c ni o’sish tartibida joylashtring. A) a < b < c B) b < a < c C) b < c < a D) c < b < a Yechish: Berilgan sonlarga teskari sonlarni taqqos- laymiz. 1 a = 11 5 = 2 1 5 , 1 b = 7 3 = 2 1 3 , 1 c = 13 6 = 2 1 6 . Bu kasrlarning butun qismlari (2 ga) teng. Ular- ning kasr qismlari 2-qoida bo’yicha taqqoslanadi va biz 1 c < 1 a < 1 b ekanligini olamiz. 4-qoidaga ko’ra, c > a > b ekan. Javob: b < a < c (B). 15. (98-10-7) Sonlarni o’sish tartibida joylashtiring? a = 49 150 ; b = 102 300 ; c = 22 75 A) a < c < b B) b < c < a C) c < a < b D) b < c < a 16. (98-10-53) Sonlarni o’sish tartibida joylashtiring? a = 5 11 ; b = 6 13 ; c = 7 15 A) a < b < c B) b < a < c C) b < c < a D) c < b < a 17. (99-4-10) Sonlarni kamayish tartibida joylashtir- ing? a = 7 36 ; b = 11 34 ; c = 7 32 ; d = 9 25 A) a > b > c > d B) b > a > d > c C) d > a > b > c D) d > b > c > a 18. (02-3-3) 3 7 , 4 17 , 21 23 sonlariga bo’lganda, butun son chiqadigan eng kichik natural sonni toping. A) 84 B) 36 C) 42 D) 63 Yechish: 3 7 , 4 17 , 21 23 sonlariga bo’lganda, butun son chiqadigan eng kichik natural sonni n bilan belgilaymiz. Demak, n : 3 7 = n · 7 3 , n : 4 17 = n · 17 4 , n : 21 23 = n · 23 21 sonlari butun bo’lishi uchun n soni 3, 4 va 21 sonlariga bo’linishi kerak. Minimallik shartidan n ularning EKUKi ekanligi kelib chiqadi. EKUK topish qoidasiga ko’ra, K(3; 4; 21) = 84. Javob: 84 (A). 19. (03-7-6) 2 7 , 4 11 , 6 13 sonlariga bo’linganda, bo’linma butun son chiqadigan eng kichik natural sonni to- ping. A) 6 B) 12 C) 18 D) 24 20. (00-5-7) 1 30 va 1 45 kasr umumiy maxrajining bar- cha natural bo’luvchilari soni nechta? A) 11 B) 7 C) 12 D) 13 21. (03-6-5) Agar 29 31 + 38 41 + 47 51 = a bo’lsa, 2 31 + 3 41 + 4 51 quyidagilardan qaysi biriga teng? A) 3 − a B) 4 − a C) 5 − a D) 3 − a 2 Yechish: Izlanayotgan sonni x bilan belgilaymiz. 29 31 + 38 41 + 47 51 = a va 2 31 + 3 41 + 4 51 = x sonlarni (chap tomoniga chap tomonini, o’ng tomoniga o’ng tomonini) qo’shamiz. Natijada 1 + 1 + 1 = = a + x ni olamiz. Bu yerdan x = 3 − a. Javob: 3 − a (A). 22. Agar 3 5 + 7 15 + 21 25 = a bo’lsa, 2 5 + 8 15 + 4 25 quyidagilardan qaysi biriga teng? A) 3 − a B) 4 − a C) 5 − a D) 3 − a 2 23. (03-7-8) Agar 29 31 + 38 41 + 47 51 + 56 61 = a bo’lsa, 2 31 + 3 41 + 4 51 + 5 61 quyidagilardan qaysi biriga teng? A) 3 − a B) 4 − a C) 5 − a D) 3 − a 2 16 24. Agar 3 10 + 5 20 + 7 30 + 9 40 = a bo’lsa, 2 10 + 5 20 + 8 30 + 11 40 quyidagilardan qaysi biriga teng? A) 3 − a B) 4 − a C) 2 − a D) 3 − 2a 25. (97-10-6) Hisoblang. 1 8 17 · 3 2 5 : 11 12 · 2 1 5 : 4 9 A) 2, 7 B) 24 3 17 C) 27 D) 29 1 9 Yechish: Aralash kasrlarni noto’g’ri kasrga kelti- rib so’ngra bo’lish amalini 1.2.1-dagi 7-qoidaga ko’ra ko’paytmaga almashtiramiz, keyin ularning surat va maxrajlarini qisqartiramiz: 1 8 17 · 3 2 5 : 11 12 · 2 1 5 : 4 9 = 25 17 · 17 5 · 12 11 · 11 5 · 9 4 = 27. Javob: 27 (C). 26. (96-7-6) Hisoblang. 5 5 7 : 2 2 5 · 5 1 4 : 1 1 6 · 2 3 A) 7 1 7 B) 8 1 7 C) 6 6 7 D) 5 5 7 27. (97-3-6) Hisoblang. 3 4 · 1 1 7 : 2 15 · 12 1 4 : 7 1 2 A) 10 1 2 B) 11 C) 9 1 4 D) 7 1 2 28. (97-7-6) Hisoblang. 42 95 · 1 3 14 : 3 5 : 2 · 4 3 4 A) 13 8 B) 1 3 8 C) 2 1 8 D) 1 5 7 29. (98-3-8) Hisoblang. 3 1 3 · 2 1 4 · ³ − 1 2 ´ · 4 5 A) 3 B) −3 C) 2, 5 D) −2, 5 30. (98-10-56) Hisoblang. 2 2 3 : 1 1 7 · 3 3 7 · ³ − 1 4 ´ A) 4 B) 3 C) −2 D) 2 7 31. (07-108-1) Hisoblang. 15 56 · 1 1 7 : 2 15 · 24 1 2 : 7 1 2 A) 11 B) 10 1 2 C) 7 1 2 D) 21 32. (96-7-9) Hisoblang. ³ 7 1 3 − 6 7 8 ´ : 3 4 + 8 8 9 · 2 1 80 A) 17 2 3 B) 18 1 2 C) 21 1 2 D) 16 1 3 Yechish: Dastlab qavs ichidagi ayirmani hisoblay- miz. 7 1 3 − 6 7 8 = 7 − 6 + 1 3 − 7 8 = 1 + 8 · 1 − 3 · 7 24 = = 1 − 13 24 = 24 − 13 24 = 11 24 . Aralash kasrlarni noto’g’ri kasrlarga aylantirib, amallarni bajaramiz: 11 24 · 4 3 + 80 9 · 161 80 = 11 18 + 161 9 = 11 + 322 18 = 333 18 = 37 2 = 18 1 2 . Javob: 18 1 2 (B). 33. (97-1-3) Hisoblang. 1 1 4 + 5 12 : ( 1 3 · 2 1 2 − 7 8 ) A) 11 1 4 B) Download 1.09 Mb. Do'stlaringiz bilan baham: |
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