Odpowiedź:
10.
a)
(4x + 5)/2x = 4
założenie:
2x ≠ 0
x ≠ 0
D: x ∈ R \ {0}
(4x + 5)/2x = 4
4x + 5 = 2x * 4
4x + 5 = 8x
4x - 8x = - 5
- 3x = - 5
3x = 5
x = 5/3 = 1 2/3
b)
15/x = x - 2
założenie:
x ≠ 0
D: x ∈ R \ {0}
15/x = x - 2
15 = x(x - 2)
15 = x² - 2x
x² - 2x - 15 = 0
a = 1 , b = - 2 , c = - 15
Δ = b² - 4ac = (- 2)² - 4 * 1 * (- 15) = 4 + 60 = 64
√Δ = √64 = 8
x₁ = ( - b - √Δ)/2a = (2 - 8)/2 = - 6/2 = 3
x₂ = (- b + √Δ)/2a = (2 + 8)/2 = 10/2 = 5
c)
x = (7x - 4)/(2x - 2)
założenie:
2x - 2 ≠ 0
2x ≠ 2
x ≠ 2/2
x ≠ 1
D: x ∈ R \ {1 }
x = (7x - 4)/(2x - 2)
x(2x - 2) = 7x - 4)
2x² - 2x = 7x - 4
2x² - 2x - 7x + 4 = 0
2x² - 9x + 4 = 0
a = 2 , b = - 9 , c = 4
Δ = b² - 4ac = (- 9)² - 4 * 2 * 4 = 81 - 32 = 49
√Δ = √49 = 7
x₁ = ( - b - √Δ)/2a = ( 9 - 7)/4 = 2/4 = 1/2
x₂ = (- b + √Δ)/2a = (9 + 7)/4 = 16/4 = 4
d)
x + 2 = (2 - 2x)/x
założenie:
x ≠ 0
D: x ∈ R \ {0}
x + 2 = (2 - 2x)/x
x(x + 2) = 2 - 2x
x² + 2x = 2 - 2x
x² + 2x + 2x - 2 = 0
x² + 4x - 2 = 0
a = 1 , b = 4 , c = - 2
Δ = b² - 4ac = 4² - 4 * 1 * (- 2) = 16 + 8 = 24
√Δ = √24 = √(4 * 6) = 2√6
x₁ = ( - b - √Δ)/2a = (- 4 - 2√6)/2 = - 2(2 + √6)/2 = - (2 + √6)
x₂ = (- b + √Δ)/2a = (- 4 + 2√6)/2 = 2(√6 - 2)/2 = √6 - 2
e)
(2x + 7)/(x + 1) = x - 1
założenie:
x + 1 ≠ 0
x ≠ - 1
D: x ∈ R \ {- 1}
(2x + 7)/(x + 1) = x - 1)
2x + 7 = (x - 1 )(x + 1)
2x + 7 = x² - 1
x² - 1 - 2x - 7 = 0
x² - 2x - 8 = 0
a = 1 , b = - 2 , c = - 8
Δ = b² - 4ac = (- 2)² - 4 * 1 * (- 8) = 4 + 32 = 36
√Δ = √36 = 6
x₁ = ( - b - √Δ)/2a = (2 - 6)/2 = - 4/2 = - 2
x₂ = (- b + √Δ)/2a = (2 + 6)/2 = 8/2 = 4
f)
(6x + 2)/(- 1 - 3x) = - 2x - 2
założenie:
- 1 - 3x ≠ 0
- 3x ≠ 1
3x ≠ - 1
x ≠ - 1/3
D: x ∈ R \ { - 1/3 }
(6x + 2)/(- 1 - 3x) = - 2x - 2
6x + 2 = (- 2x - 2)(- 1 - 3x)
6x + 2 = 2x + 2 + 6x² + 6x
6x + 2 = 6x² + 8x + 2
6x² + 8x - 6x + 2 - 2 = 0
6x² + 2x = 0
2x(3x + 1) = 0
2x = 0 ∨ 3x + 1 = 0
x = 0 ∨ 3x = - 1
x = 0 ∨ x = - 1/3
Ponieważ x = - 1/3 nie należy do dziedziny więc :
x = 0
