Question

12. Arătați ca A inclus B, unde : ​

Answer 1

Explicație pas cu pas:

c) 2⁰=1, 2¹=2, 2²=4, 2³=8

A = {1, 2, 4, 8}

x îl divide pe 8, x ∈ N

B = {1, 2, 4, 8}

=> A ⊂ B

d) |x + 3| < 5 <=> -5 < x + 3 < 5

-8 < x < 2, x ∈ Z

A = {-7, -6, -5, -4, -3, -2, -1, 0, 1}

|x| ≤ 7 <=> -7 ≤ x ≤ 7, x ∈ Z

B = {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7}

=> A ⊂ B

e) 4x + 11 > 67 <=> 4x > 67 - 11

4x > 56 <=> x > 56:4 => x > 14, x ∈ N

A = {15, 16, 17, ...}

7x - 5 ≥ 72 <=> 7x ≥ 72 + 5

7x ≥ 77 <=> x ≥ 77:7 => x ≥ 11, x ∈ N

B = {11, 12, 13, ...}

=> A ⊂ B

f) (x² - 3) | 22 ; x² ≥ 0 => x² - 3 ≥ -3

(x² - 3) este divizor al lui 22, x ∈ N

x² ≥ 0 <=> x² - 3 ≥ - 3

=> (x² - 3) ∈ {-2, -1, 1, 2, 11, 22}

x² - 3 = -2 => x² = 1 => x = -1 ∉ N, x = 1 ∈ N

x² - 3 = -1 => x² = 2 => x ∉ N

x² - 3 = 1 => x² = 4 => x = -2 ∉ N, x = 2 ∈ N

x² - 3 = 2 => x² = 5 => x ∉ N

x² - 3 = 11 => x² = 14 => x ∉ N

x² - 3 = 22 => x² = 25 => x = -5 ∉ N, x = 5 ∈ N

A = {1, 2, 5}

|x| ≤ 5 <=> -5 ≤ x ≤ 5

B = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

=> A ⊂ B