Question

aratati ca A ( B, unde:​

Answer 1

Răspuns:

A ⊂ B

Explicație pas cu pas:

b) A = {2, 3, 6, 9}

54 : x <=> B = {1, 2, 3, 6, 9, 18, 27, 54}

=> A ⊂ B

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c) 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8

A = {1, 2, 4, 8}

x | 8 <=> B = {1, 2, 4, 8}

=> A ⊂ B

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

- 5 - 3 < x < 5 - 3 <=> - 8 < x < 2

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

|x| ≤ 7 <=> -7 ≤ x ≤ 7

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

=> A ⊂ B

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e) 4x + 11 > 67

4x > 56 => x > 14

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

7x - 5 ≥ 72 <=> 7x ≥ 77 => x ≥ 11

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

=> A ⊂ B

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f) (x² - 3) | 22 => (x² - 3) ∈ {-22, -11, -2, -1, 1, 2, 11, 22}

x² ≥ 0 => x² - 3 ≥ - 3

(x² - 3) ∈ {-2, -1, 1, 2, 11, 22}

x² - 3 = -2 <=> x² = 1 = 1² => x = ±1 => x = 1

x² - 3 = -1 <=> x² = 2 => x = ± √2 ∉ N

x² - 3 = 1 <=> x² = 4 = 2² => x = ±2 => x = 2

x² - 3 = 2 <=> x² = 5 => x = ± √5 ∉ N

x² - 3 = 11 <=> x² = 14 => x = ± √14 ∉ N

x² - 3 = 22 <=> x² = 25 => x = ± 5 => x = 5

A = {1, 2, 5}

|x| ≤ 5 <=> -5 ≤ x ≤ 5

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

=> A ⊂ B