Question

Matematyka, cz. 1
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Answer 1

Odpowiedź:

zad 1

a)

[(4x - 3)(x - 2)]/(6x - 2) = 0

założenie:

6x - 2 ≠ 0

2(3x - 1) ≠ 0

3x - 1 ≠ 0

3x ≠ 1

x ≠ 1/3

D: x ∈ R \ {1/3}

(4x - 3)(x - 2) = 0

4x - 3 = 0 ∨ x - 2 = 0

4x = 3 ∨ x = 2

x = 3/4 ∨ x = 2

b)

4/(x + 3) = x + 3

założenie:

x + 3 ≠ 0

x ≠ - 3

D: x ∈ R \ {3}

4/(x + 3) = x + 3 | * (x + 3)

4 = (x + 3)(x + 3) = (x + 3)² = x² + 6x + 9

x² + 6x + 9 = 4

x² + 6x + 9 - 4 = 0

x² + 6x + 5 = 0

a = 1 , b = 6 , c = 5

Δ = b² - 4ac = 6² - 4 * 1 * 5 = 36 - 20 = 16

√Δ = √16 = 4

x₁ = (- b - √Δ)/2a = (- 6 - 4)/2 = - 10/2 = - 5

x₂ = (- b + √Δ)/2a = (- 6 + 4)/2 = - 2/2 = - 1

zad 2

a)

Ix + 6I = 2

x + 6 = 2 ∨ x + 6 = - 2

x = 2 - 6 ∨ x = - 2 - 6

x = - 4 ∨ x = - 8

b)

Ix - 3I = 7

x - 3 = 7 ∨ x - 3 = - 7

x = 7 + 3 ∨ x = - 7 + 3

x = 10 ∨ x = - 4

zad 3

a)

Ix - 6I ≤ 2

x - 6 ≤ 2 ∧ x - 6 ≥ - 2

x ≤ 2 + 6 ∧ x ≥ - 2 + 6

x ≤ 8 ∧ x ≥ 4

x ∈ < 4 , 8 >

b)

Ix - 3I ≥ 5

x - 3 ≥ 5 ∨ x - 3 ≤ - 5

x ≥ 5 + 3 ∧ x ≤ - 5 + 3

x ≥ 8 ∧ x ≤ - 2

x ∈ ( - ∞ , - 2 > ∪ < 8 , + ∞ )

∧ - znaczy " i "

∨ - znaczy " lub "