§. Natural va butun sonlar

- §. 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.

1. 
   
   
2. 
   
   
3. 
   -
   
4. 
   (b ) tengsizlik -b tengsizlikka ekvivalent hisoblanadi.
   
5. 
   
   
6. 
   
   
7. 
   
   
8. 
   Max(a;b) =
   
9. 
   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)
6. ifodanisoddalashtiring. ( xϵ(– ∞; 1) )
A) x² +x+1 B) – (x²+x+1) C)x² +x D)x²– x+1
40. ifodanisoddalashtiring.
( xϵ(1; 3) )
A) x² +x+1 B) –(x² +x+1) C)x² +x D)x²– x+1
41. ifodanisoddalashtiring.( xϵ(3; ∞) )
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 C) c<0 D) b
79. funksiyaning eng katta qiymatini toping
A) 9 B) 12 C) 15 D) 24

12.