Natural va butun sonlar
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§. Natural va butun sonlar-hozir.org
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- 46. Quyidagi tengsizliklarni doimo to’g’riligini ko’rsating. 1) (a>0) 2) ) 3. 4.
- 15. a 2 b 2 +b 2 c 2 +a 2 c 2 abc(a+b+c) 16. 2(a 3 +b 3 +c 3 ) ab(a+b)+ac(a+c)+bc(b+c)
- +2x+6y+4>0 20. (n>1) 21.
- 48. Agar x>0, y>0, z>0 va x+y+z =12 shartlar o’rinli bo’lsa, U = x 2 yz 3
(1-x1)2+(x1 -x2)2+(x2 -x3)2+……+(xn-1 –xn)2+xn2 = tenglamani yagona yechimi borligini ko’rsating. Hamda bu yechimlarni o’zini toping.
45. Shunday nN sonini topingki tenglamalar sistemasi qanoatlantiruvchi x1, x2 …….xn musbat sonlar mavjud bo’lsin. Topilgan n natural soni bo’yicha sistemani yeching. 46. Quyidagi tengsizliklarni doimo to’g’riligini ko’rsating. 1) (a>0) 2) ) 3. 4. 5. (2n)! < (n(n+1))n 6. ( 7. Shlyomilx masalasi. n>2 uchun quyidagi tengsizlikni doimo o’rinli ekanligiligi ko’rsating. 8. 9. 10. 11. ab2c3d4 12. x2+y2+z2xy+yz+xz 13. x2+y2+1 xy+x+y 14. x4+y4+88xy 15. a2b2+b2c2+a2c2abc(a+b+c) 16. 2(a3+b3+c3) ab(a+b)+ac(a+c)+bc(b+c) 17. x2 -5xy+7y2 0 18. 4xy –3x2 -8y20 19. x2 +2xy+3y2+2x+6y+4>0 20. (n>1)
28. 29. nn>1 ( 30. 47. tengsizliklar sistemasini yeching.
48. Agar x>0, y>0, z>0 va x+y+z =12 shartlar o’rinli bo’lsa, U = x2yz3 ifodaning eng katta qiymatini toping. A)6912 B)6000 C) 2000 D)aniqlab bo’lmaydi
- §. Modulli ifodalar x haqiqiy sonning absolyut qiymati (moduli) deb nomanfiy bo’lgan ko’rinishdagi songa aytiladi. Uning umumiy ko’rinishi quyidagicha bo’ladi. ning geometrik ma’nosi koordinata boshidan x gacha bo’lgan masofani anglatadi. Absolyut qiymatlar bo’yicha quyidagi munosabatlar o’rinlidir. - (b ) tengsizlik -b tengsizlikka ekvivalent hisoblanadi. Max(a;b) = min(a;b) = 1. ifodaningqiymatinimodulbelgisisizyozing. A)1 B)2 C)3 D)4 2. ifodaningqiymatinimodulbelgisisizyozing. A)1B)2 C)3 D)4 3. ifodaningqiymatinimodulbelgisisizyozing. A)1B)2C)3 D)4 4. ifodaningqiymatinimodulbelgisisizyozing. A)1B)2 C)3 D)4 5. Ifodaning qiymatini modul belgisisiz yozing. A) B) C) D)
A) x2 +x+1B) – (x2 +x+1) C)x2 +x D)x2– x+1 42. ifodanisoddalashtiring. ( aϵ(– ∞; – 2) ) A)– a/2 B) (a – 1)/2 C) a(a – 1)/2 D) a(a + 1)/2 43. ifodanisoddalashtiring.( aϵ(– 2; ∞) ) A) – a/2 B) (a – 1)/2C) a(a – 1)/2 D) a(a + 1)/2 44. ifodanisoddalashtiring. ( mϵ(– ∞; – 2) ) A) 1/(m + 2) B) – 1/(m + 2) C) 1/(m – 2) D)– 1/(m – 2) 45. ifodanisoddalashtiring. ( mϵ(– 2; 0) ) A)1/(m + 2) B) – 1/(m + 2) C) 1/(m – 2) D) – 1/(m – 2) 46. ifodanisoddalashtiring. ( mϵ(3; ∞) ) A) 1/(m + 2) B) – 1/(m + 2) C) 1/(m – 2) D) – 1/(m – 2) 47. ifodanisoddalashtiring. ( mϵ(0; 3) ) A) 1/(m + 2) B) – 1/(m + 2) C) 1/(m – 2) D) – 1/(m – 2) 48. ifodanisoddalashtiring. ( xϵ(– ∞; – 1) ) A) x– 2 B)(x2 + 4)/(x – 2) C) –(x+2) D) x+2 49. ifodanisoddalashtiring.( xϵ(– 1; 1) ) A) x– 2B)(x2 + 4)/(x – 2) C) –(x+2) D) x+2 50. ifodanisoddalashtiring. ( xϵ(1; 2) ) A) x– 2 B)(x2 + 4)/(x – 2)C) –(x+2) D) x+2 51. ifodanisoddalashtiring. ( xϵ(2; ∞) ) A) x– 2 B)(x2 + 4)/(x – 2) C) –(x+2) D) x+2 52. ifodanisoddalashtiring. ( xϵ(– ∞; – 1) ) A)x/(x – 1) B) x/(1 – x) C) – x/(x + 1) D) x/(x + 1) 53. ifodanisoddalashtiring. ( xϵ(– 1; 0) ) A) x/(x – 1) B) x/(1 – x) C) – x/(x + 1) D) x/(x + 1) 54. ifodanisoddalashtiring. ( xϵ[0; 1) ) A) x/(x – 1) B) x/(1 – x) C) – x/(x + 1) D) x/(x + 1) 55. ifodanisoddalashtiring. ( xϵ(1; ∞) ) A) x/(x – 1) B) x/(1 – x) C) – x/(x + 1)D) x/(x + 1) 56. ifodanisoddalashtiring. ( xϵ(– ∞; 2) ) A) x2 – 4x–12 B) (x+2)2 C) x2 – 4x+12 D)x2 + 4x+12 57. ifodanisoddalashtiring. ( xϵ(2; ∞) ) A) x2 – 4x–12B) (x+2)2 C) x2 – 4x+12 D)x2 + 4x+12 58. ifodanisoddalashtiring. ( xϵ(–∞; 0) ) A)– 1/x B) 1/x C)x D)–x 59. ifodanisoddalashtiring.( xϵ(0; 1) ) A) – 1/x B) 1/x C)x D)– x 60. ifodanisoddalashtiring.( xϵ(1; 2) ) A)– 1/x B) 1/x C)x D)– x 61. ifodanisoddalashtiring. ( xϵ(2; 3) ) A)– 1/xB) 1/x C)x D)– x 62. ifodanisoddalashtiring.( xϵ(3; ∞) ) A)– 1/xB) 1/x C)x D)– x 63. ifodanisoddalashtiring.( xϵ( – ∞; – 3) ) A) 1/(a + 1) B) 1/(a + 3) C) 1/(a – 1) D) 1/(a – 3) 64. ifodanisoddalashtiring. ( xϵ(– 3; – 1) ) A) 1/(a + 1) B) 1/(a + 3) C) 1/(a – 1) D) 1/(a – 3) 65. ifodanisoddalashtiring. ( xϵ(–1; 2) ) A) 1/(a + 1) B) 1/(a + 3) C) 1/(a – 1) D) 1/(a – 3) 66. ifodanisoddalashtiring.( xϵ(2; ∞) ) A) 1/(a + 1) B) 1/(a + 3) C) 1/(a – 1) D) 1/(a – 3) 67. ifodanisoddalashtiring. ( xϵ(– ∞; 0) ) A) B) C) D) 68. ifodanisoddalashtiring.( xϵ(0; 1) ) A) B) C) D) 69. ifodanisoddalashtiring. ( xϵ[ 1; ∞) ) A) B) C) D) 70. ifodanisoddalashtiring.( xϵ(– ∞; 0) ) A) B) C) D) 71. ifodanisoddalashtiring.( xϵ[ 0; 1/3) ) A) B) C) D) 72. ifodanisoddalashtiring.( xϵ(1/3; 1) ) A) B) C) D) 73. ifodanisoddalashtiring. ( xϵ(1; ∞) ) A) B) C) D) 74. ifodanisoddalashtiring.( xϵ( – ∞; – 3/2) ) A) B) C) D) 75. ifodanisoddalashtiring.( xϵ(– 3/2; 0) ) A) B) C) D) 76. ifodanisoddalashtiring.( xϵ(0; 3) ) A) B) C) D) 77. ifodanisoddalashtiring.( xϵ(3; ∞) ) A) B) C) D) 78. |−abc| = −abc, |a−b| = −b + a va |−b| = b bo’lsa, quyidagilardan qaysi biri har doim o’rinli. A) b<0 B) 0 79. funksiyaning eng katta qiymatini toping A) 9 B) 12 C) 15 D) 24 Download 5.8 Mb. Do'stlaringiz bilan baham: |
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